Nelson Functions 11
1st Edition
ISBN: 9780176332037
Textbook solutions
All Solutions
Page 468: Practice Questions
Exercise 1
Step 1
1 of 5
[begin{gathered}
{mathbf{Identifying}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{A sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{is arithmetic if the 1st difference between consecutive terms is constant,}} hfill \
hfill \
{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d hfill \
hfill \
{mathbf{General}},{mathbf{Term}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{The }}{{text{n}}^{th}};{text{term of an arithmetic sequence }} hfill \
{text{with first term }}{t_1},{text{and common difference }}d{text{ is}} hfill \
hfill \
{t_n} = {t_1} + left( {n – 1} right)d hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,, hfill \
{t_1},,,,,{t_n} = {t_{n – 1}} + d,,,,,{text{where }}n > 1,,, hfill \
hfill \
hfill \
end{gathered} ]
{mathbf{Identifying}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{A sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{is arithmetic if the 1st difference between consecutive terms is constant,}} hfill \
hfill \
{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d hfill \
hfill \
{mathbf{General}},{mathbf{Term}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{The }}{{text{n}}^{th}};{text{term of an arithmetic sequence }} hfill \
{text{with first term }}{t_1},{text{and common difference }}d{text{ is}} hfill \
hfill \
{t_n} = {t_1} + left( {n – 1} right)d hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,, hfill \
{t_1},,,,,{t_n} = {t_{n – 1}} + d,,,,,{text{where }}n > 1,,, hfill \
hfill \
hfill \
end{gathered} ]
Step 2
2 of 5
a.) This is an arithmetic sequence with $t_1=2$ and $d=8-2=6$.
Step 3
3 of 5
b.)
General Term: $t_n=2+(n-1)(6)=6n-4$
Recursive Formula: $t_1=2$ ; $t_n=t_{n-1}+6$ where $n>1$
Step 4
4 of 5
c.)
Result
5 of 5
a.) arithmetic
b.) $t_n=6n-4$
c.) see graph
Exercise 2
Step 1
1 of 1
A sequence is arithmetic if the difference between two consecutive terms is constant. The difference is also known as the common difference.
Exercise 3
Step 1
1 of 5
[begin{gathered}
{mathbf{Identifying}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{A sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{is arithmetic if the 1st difference between consecutive terms is constant,}} hfill \
hfill \
{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d hfill \
hfill \
{mathbf{General}},{mathbf{Term}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{The }}{{text{n}}^{th}};{text{term of an arithmetic sequence }} hfill \
{text{with first term }}{t_1},{text{and common difference }}d{text{ is}} hfill \
hfill \
{t_n} = {t_1} + left( {n – 1} right)d hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,, hfill \
{t_1},,,,,{t_n} = {t_{n – 1}} + d,,,,,{text{where }}n > 1,,, hfill \
hfill \
hfill \
end{gathered} ]
{mathbf{Identifying}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{A sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{is arithmetic if the 1st difference between consecutive terms is constant,}} hfill \
hfill \
{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d hfill \
hfill \
{mathbf{General}},{mathbf{Term}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{The }}{{text{n}}^{th}};{text{term of an arithmetic sequence }} hfill \
{text{with first term }}{t_1},{text{and common difference }}d{text{ is}} hfill \
hfill \
{t_n} = {t_1} + left( {n – 1} right)d hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,, hfill \
{t_1},,,,,{t_n} = {t_{n – 1}} + d,,,,,{text{where }}n > 1,,, hfill \
hfill \
hfill \
end{gathered} ]
Step 2
2 of 5
a.) This is an arithmetic sequence with $t_1=58$ and $d=73-58=15$.
General Term: $t_n=58+(n-1)(15)=15n+43$
Recursive Formula: $t_1=58$ ; $t_n=t_{n-1}+15$ where $n>1$
Step 3
3 of 5
b.) This is an arithmetic sequence with $t_1=-49$ and $d=(-40)-(-49)=9$.
General Term: $t_n=-49+(n-1)(9)=9n-58$
Recursive Formula: $t_1=-49$ ; $t_n=t_{n-1}+9$ where $n>1$
Step 4
4 of 5
c.) This is an arithmetic sequence with $t_1=81$ and $d=75-81=-6$.
General Term: $t_n=81+(n-1)(-6)=87-6n$
Recursive Formula: $t_1=81$ ; $t_n=t_{n-1}-6$ where $n>1$
Result
5 of 5
a.) $t_n=15n+43$
b.) $t_n=9n-58$
c.) $t_n=87-6n$
Exercise 4
Step 1
1 of 3
If $t_j$ and $t_k$ are terms in the arithmetic sequence, then
$t_k-t_j=(k-j)d$
Here, we’re given with $t_7=465$ and $t_{13}=219$
$$
219-465=(13-7)(d)implies d=-41
$$
Step 2
2 of 3
Using the calculated $d$, the 100th term is
$$
t_{100}-465=(100-7)(-41)implies t_{100}=-3348
$$
Result
3 of 3
$$
t_{100}=-3348
$$
t_{100}=-3348
$$
Exercise 5
Step 1
1 of 4
[begin{gathered}
{mathbf{Identifying}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{A sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{is arithmetic if the 1st difference between consecutive terms is constant,}} hfill \
hfill \
{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d hfill \
hfill \
{mathbf{General}},{mathbf{Term}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{The }}{{text{n}}^{th}};{text{term of an arithmetic sequence }} hfill \
{text{with first term }}{t_1},{text{and common difference }}d{text{ is}} hfill \
hfill \
{t_n} = {t_1} + left( {n – 1} right)d hfill \
end{gathered} ]
{mathbf{Identifying}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{A sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{is arithmetic if the 1st difference between consecutive terms is constant,}} hfill \
hfill \
{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d hfill \
hfill \
{mathbf{General}},{mathbf{Term}},{mathbf{of}},{mathbf{Arithmetic}},{mathbf{Sequence}}:,,{text{The }}{{text{n}}^{th}};{text{term of an arithmetic sequence }} hfill \
{text{with first term }}{t_1},{text{and common difference }}d{text{ is}} hfill \
hfill \
{t_n} = {t_1} + left( {n – 1} right)d hfill \
end{gathered} ]
Step 2
2 of 4
The height of the plant increases constantly every week by $d=20-7$ mm and is initially at 7 mm on the first week. Thus, we can model the height of the plant as a function of week number by an arithmetic sequence with $t_1=7$ and $d=13$.
Step 3
3 of 4
The general term is
$t_n=7+(n-1)(13)implies t_n=13n-6$
We want to know the week number such that the height is more than $100$ mm.
$13n-6>100$
$n>dfrac{100+6}{13}$
$n>8.1538$
Thus, the plant will be at least $100$ mm high on the $boxed{bold{9^{th}; week}}$
Result
4 of 4
9$^{th}$ week
Exercise 6
Step 1
1 of 1
A sequence is geometric if the ratio between two consecutive terms is constant.
Exercise 7
Step 1
1 of 8
[begin{gathered}
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
Step 2
2 of 8
a.) This is a geometric sequence with $t_1=5$ and $r=dfrac{15}{5}=3$
General Term: $t_n=5cdot 3^{n-1}$
Sixth term: $t_6=5cdot 3^{6-1}=1215$
Step 3
3 of 8
b.) No common difference and no common ratio in the consecutive terms, thus, it is neither arithmetic nor geometric.
Step 4
4 of 8
c.) This is a geometric sequence with $t_1=288$ and $r=dfrac{14.4}{288}=0.05$.
General Term: $t_n=288cdot 0.05^{n-1}$
Sixth term: $t_6=288cdot 0.05^{6-1}=0.00009$
Step 5
5 of 8
d.) This is an arithmetic sequence with $t_1=10$ and $d=50-10=40$.
General Term: $t_n=10+(n-1)(40)=40n-30$
Sixth term: $t_6=40(6)-30=210$
Step 6
6 of 8
e.) This is an arithmetic sequence with $t_1=19$ and $d=10-19=-9$.
General Term: $t_n=19+(-9)(n-1)=28-9n$
Sixth term: $t_6=28-9(6)=-26$
Step 7
7 of 8
f.) This is a geometric sequence with $t_1=512$ and $r=dfrac{384}{512}=0.75$.
General Term: $t_n=512cdot 0.75^{n-1}$
Sixth term: $t_6=512cdot 0.75^{6-1}=121.5$
Result
8 of 8
a.) $t_n=5cdot 3^{n-1}$
b.) neither arithmetic nor geometric
c.) $t_n=288cdot 0.05^{n-1}$
d.) $t_n=40n-30$
e.) $t_n=28-9n$
f.) $t_n=512left(dfrac{3}{4}right)^{n-1}$
Exercise 8
Step 1
1 of 5
[begin{gathered}
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
Step 2
2 of 5
a.) This is a geometric sequence with $t_1=7$ and $r=-3$
Recursive Formula: $t_1=7$ , $t_n=-3t_{n-1}$, where $n>1$
General Term: $t_n=7cdot (-3)^{n-1}$
$t_1=7$
$t_2=7(-3)^{2-1}=-21$
$t_3=7(-3)^{3-1}=63$
$t_4=7(-3)^{4-1}=-189$
$t_5=7(-3)^{5-1}=567$
Step 3
3 of 5
a.) This is a geometric sequence with $t_1=12$ and $r=0.5$
Recursive Formula: $t_1=12$ , $t_n=-dfrac{1}{2}cdot t_{n-1}$, where $n>1$
General Term: $t_n=12cdot left(dfrac{1}{2}right)^{n-1}$
$t_1=12$
$t_2=12(0.5)^1=6$
$t_3=12(0.5)^2=3$
$t_4=12(0.5)^3=dfrac{3}{2}$
$t_5=12(0.5)^4=dfrac{3}{4}$
Step 4
4 of 5
c.) The common ratio is $r=dfrac{144}{36}=4$.
With $t_2=36$, we can get $r=dfrac{t_2}{t_1}implies t_1=36/4=9$
Recursive Formula: $t_1=12$ , $t_n=-dfrac{1}{2}cdot t_{n-1}$, where $n>1$
General Term: $t_n=9cdot left(4right)^{n-1}$
$t_1=9$
$t_2=9(4)^1=36$
$t_3=9(4)^2=144$
$t_4=9(4)^3=576$
$t_5=9(4)^4=2304$
Result
5 of 5
a.) $7$, $-21$, $63$, $-189$, $567$
b.) $12$, $6$, $3$, $dfrac{3}{2}$, $dfrac{3}{4}$
c.) $9$, $36$, $144$, $576$, $2304$
Exercise 9
Step 1
1 of 6
[begin{gathered}
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
Step 2
2 of 6
a.) This is an arithmetic sequence with $t_1=9$ and $d=5$
The first five terms are
$t_1=4(1)+5=9$
$t_2=4(2)+5=13$
$t_3=4(3)+5=17$
$t_4=4(4)+5=21$
$t_5=4(5)+5=25$
Step 3
3 of 6
[begin{gathered}
{text{b}}{text{.)}},{text{We shall write the first 5 terms }} hfill \
{t_n} = frac{1}{{7n – 3}} hfill \
{t_1} = frac{1}{{7left( 1 right) – 3}} = frac{1}{4} hfill \
{t_2} = frac{1}{{7left( 2 right) – 3}} = frac{1}{{11}} hfill \
{t_3} = frac{1}{{7left( 3 right) – 3}} = frac{1}{{18}} hfill \
{t_4} = frac{1}{{7left( 4 right) – 3}} = frac{1}{{25}} hfill \
{t_5} = frac{1}{{7left( 5 right) – 3}} = frac{1}{{32}} hfill \
{text{Let us take the first difference of consecutive terms}} hfill \
{t_2} – {t_1} = frac{1}{{11}} – frac{1}{4} = – frac{7}{{44}} hfill \
{t_3} – {t_2} = frac{1}{{18}} – frac{1}{{11}} = – frac{7}{{198}} hfill \
{text{The first difference of consecutive terms is not constant,}} hfill \
{text{thus it is NOT arithmetic}} hfill \
{text{Let’s try taking the ratio of consecutive terms}} hfill \
frac{{{t_2}}}{{{t_1}}} = frac{{1/11}}{{1/4}} = frac{4}{{11}} hfill \
frac{{{t_3}}}{{{t_2}}} = frac{{1/18}}{{1/11}} = frac{{11}}{{18}} hfill \
{text{The ratios of consecutive terms is not constant,}} hfill \
{text{thus it is}}{mathbf{ ;neither }},{mathbf{;arithmetic }},{mathbf{;nor}},{mathbf{ ;geometric}}{mathbf{.}} hfill \
hfill \
end{gathered} ]
{text{b}}{text{.)}},{text{We shall write the first 5 terms }} hfill \
{t_n} = frac{1}{{7n – 3}} hfill \
{t_1} = frac{1}{{7left( 1 right) – 3}} = frac{1}{4} hfill \
{t_2} = frac{1}{{7left( 2 right) – 3}} = frac{1}{{11}} hfill \
{t_3} = frac{1}{{7left( 3 right) – 3}} = frac{1}{{18}} hfill \
{t_4} = frac{1}{{7left( 4 right) – 3}} = frac{1}{{25}} hfill \
{t_5} = frac{1}{{7left( 5 right) – 3}} = frac{1}{{32}} hfill \
{text{Let us take the first difference of consecutive terms}} hfill \
{t_2} – {t_1} = frac{1}{{11}} – frac{1}{4} = – frac{7}{{44}} hfill \
{t_3} – {t_2} = frac{1}{{18}} – frac{1}{{11}} = – frac{7}{{198}} hfill \
{text{The first difference of consecutive terms is not constant,}} hfill \
{text{thus it is NOT arithmetic}} hfill \
{text{Let’s try taking the ratio of consecutive terms}} hfill \
frac{{{t_2}}}{{{t_1}}} = frac{{1/11}}{{1/4}} = frac{4}{{11}} hfill \
frac{{{t_3}}}{{{t_2}}} = frac{{1/18}}{{1/11}} = frac{{11}}{{18}} hfill \
{text{The ratios of consecutive terms is not constant,}} hfill \
{text{thus it is}}{mathbf{ ;neither }},{mathbf{;arithmetic }},{mathbf{;nor}},{mathbf{ ;geometric}}{mathbf{.}} hfill \
hfill \
end{gathered} ]
Step 4
4 of 6
[begin{gathered}
{text{c}}{text{.)}},{text{We shall write the first 5 terms }} hfill \
{t_n} = {n^2} – 1 hfill \
{t_1} = {1^2} – 1 = 0 hfill \
{t_2} = {2^2} – 1 = 3 hfill \
{t_3} = {3^2} – 1 = 8 hfill \
{t_4} = ,{4^2} – 1 = 15 hfill \
{t_5} = ,{5^2} – 1 = 24 hfill \
hfill \
{text{Let us take the first difference of consecutive terms}} hfill \
{t_2} – {t_1} = 3 – 0 = 3 hfill \
{t_3} – {t_2} = 8 – 3 = 5 hfill \
{text{The first difference of consecutive terms is not constant,}} hfill \
{text{thus it is NOT arithmetic}} hfill \
{text{Let’s try taking the ratio of consecutive terms}} hfill \
frac{{{t_3}}}{{{t_2}}} = frac{8}{3} hfill \
frac{{{t_4}}}{{{t_3}}} = frac{{15}}{8} hfill \
{text{The ratios of consecutive terms is not constant,}} hfill \
{text{thus it is}}{mathbf{ ;neither }},{mathbf{;arithmetic }},{;mathbf{nor}},{mathbf{ ;geometric}}{mathbf{.}} hfill \
hfill \
end{gathered} ]
{text{c}}{text{.)}},{text{We shall write the first 5 terms }} hfill \
{t_n} = {n^2} – 1 hfill \
{t_1} = {1^2} – 1 = 0 hfill \
{t_2} = {2^2} – 1 = 3 hfill \
{t_3} = {3^2} – 1 = 8 hfill \
{t_4} = ,{4^2} – 1 = 15 hfill \
{t_5} = ,{5^2} – 1 = 24 hfill \
hfill \
{text{Let us take the first difference of consecutive terms}} hfill \
{t_2} – {t_1} = 3 – 0 = 3 hfill \
{t_3} – {t_2} = 8 – 3 = 5 hfill \
{text{The first difference of consecutive terms is not constant,}} hfill \
{text{thus it is NOT arithmetic}} hfill \
{text{Let’s try taking the ratio of consecutive terms}} hfill \
frac{{{t_3}}}{{{t_2}}} = frac{8}{3} hfill \
frac{{{t_4}}}{{{t_3}}} = frac{{15}}{8} hfill \
{text{The ratios of consecutive terms is not constant,}} hfill \
{text{thus it is}}{mathbf{ ;neither }},{mathbf{;arithmetic }},{;mathbf{nor}},{mathbf{ ;geometric}}{mathbf{.}} hfill \
hfill \
end{gathered} ]
Step 5
5 of 6
[begin{gathered}
{text{d}}{text{.)}},{text{We shall write the first 5 terms }} hfill \
{t_1} = – 17,,,,,{t_n} = {t_{n – 1}} + n – ,1,,,,,,n > 1 hfill \
{t_2} = – 17 + 2 – 1 = – 16 hfill \
{t_3} = – 16 + 3 – 1 = – 14 hfill \
{t_4} = – 14 + 4 – 1 = – 11 hfill \
{t_5} = , – 11 + 5 – 1 = – 7 hfill \
hfill \
{text{Let us take the first difference of consecutive terms}} hfill \
{t_2} – {t_1} = – 16 – left( { – 17} right) = 1 hfill \
{t_3} – {t_2} = – 14 – left( { – 16} right) = 2 hfill \
{text{The first difference of consecutive terms is not constant,}} hfill \
{text{thus it is NOT arithmetic}} hfill \
{text{Let’s try taking the ratio of consecutive terms}} hfill \
frac{{{t_2}}}{{{t_1}}} = frac{{16}}{{17}} hfill \
,frac{{{t_3}}}{{{t_2}}} = frac{{14}}{{16}} hfill \
{text{The ratios of consecutive terms is not constant,}} hfill \
{text{thus it is}}{mathbf{; neither }},{mathbf{;arithmetic }},{mathbf{;nor}},{mathbf{ ;geometric}}{mathbf{.}} hfill \
hfill \
end{gathered} ]
{text{d}}{text{.)}},{text{We shall write the first 5 terms }} hfill \
{t_1} = – 17,,,,,{t_n} = {t_{n – 1}} + n – ,1,,,,,,n > 1 hfill \
{t_2} = – 17 + 2 – 1 = – 16 hfill \
{t_3} = – 16 + 3 – 1 = – 14 hfill \
{t_4} = – 14 + 4 – 1 = – 11 hfill \
{t_5} = , – 11 + 5 – 1 = – 7 hfill \
hfill \
{text{Let us take the first difference of consecutive terms}} hfill \
{t_2} – {t_1} = – 16 – left( { – 17} right) = 1 hfill \
{t_3} – {t_2} = – 14 – left( { – 16} right) = 2 hfill \
{text{The first difference of consecutive terms is not constant,}} hfill \
{text{thus it is NOT arithmetic}} hfill \
{text{Let’s try taking the ratio of consecutive terms}} hfill \
frac{{{t_2}}}{{{t_1}}} = frac{{16}}{{17}} hfill \
,frac{{{t_3}}}{{{t_2}}} = frac{{14}}{{16}} hfill \
{text{The ratios of consecutive terms is not constant,}} hfill \
{text{thus it is}}{mathbf{; neither }},{mathbf{;arithmetic }},{mathbf{;nor}},{mathbf{ ;geometric}}{mathbf{.}} hfill \
hfill \
end{gathered} ]
Result
6 of 6
a.) arithmetic
b.) neither arithmetic nor geometric
c.) neither arithmetic nor geometric
d.) neither arithmetic nor geometric
Exercise 10
Step 1
1 of 3
[begin{gathered}
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{The number of bacteria doubles every hour and at 1 p}}{text{.m}}{text{., there were}} hfill \
{text{23 000 bacteria}}{text{. This can be modelled by a geometric sequence with}} hfill \
{t_1} = 23,000{text{ and }}r = 2.,,{text{The number of hours that elapsed from 1 p}}{text{.m}}{text{.}} hfill \
{text{to midnight is }}24 – 13 = ,11,{text{hours}} hfill \
hfill \
{text{The number of bacteria at midnight is therefore}} hfill \
hfill \
{t_{11}} = left( {23,000} right){left( 2 right)^{11}} = boxed{47,104,000,{text{bacteria}}} hfill \
end{gathered} ]
{text{The number of bacteria doubles every hour and at 1 p}}{text{.m}}{text{., there were}} hfill \
{text{23 000 bacteria}}{text{. This can be modelled by a geometric sequence with}} hfill \
{t_1} = 23,000{text{ and }}r = 2.,,{text{The number of hours that elapsed from 1 p}}{text{.m}}{text{.}} hfill \
{text{to midnight is }}24 – 13 = ,11,{text{hours}} hfill \
hfill \
{text{The number of bacteria at midnight is therefore}} hfill \
hfill \
{t_{11}} = left( {23,000} right){left( 2 right)^{11}} = boxed{47,104,000,{text{bacteria}}} hfill \
end{gathered} ]
Result
3 of 3
$47;104;000$ bacteria
Exercise 11
Step 1
1 of 3
[begin{gathered}
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{The value increases 10% each year starting from the initial price of 820}}{text{. }} hfill \
{text{This means the price is multiplied by 1}}{text{.1 each year so we can model this }} hfill \
{text{as a geometric sequence with }}{t_1} = 820{text{ and }}r = 1.1 hfill \
hfill \
n = 2010 – 2000 = 10 hfill \
{t_n} = 820 cdot {1.1^{10 – 1}} = ,,$ 1933.52 hfill \
hfill \
{text{The stamp would cost }}boxed{{text{$ 1933}}{text{.52}}},{text{in 2010}}{text{.}} hfill \
end{gathered} ]
{text{The value increases 10% each year starting from the initial price of 820}}{text{. }} hfill \
{text{This means the price is multiplied by 1}}{text{.1 each year so we can model this }} hfill \
{text{as a geometric sequence with }}{t_1} = 820{text{ and }}r = 1.1 hfill \
hfill \
n = 2010 – 2000 = 10 hfill \
{t_n} = 820 cdot {1.1^{10 – 1}} = ,,$ 1933.52 hfill \
hfill \
{text{The stamp would cost }}boxed{{text{$ 1933}}{text{.52}}},{text{in 2010}}{text{.}} hfill \
end{gathered} ]
Result
3 of 3
$$
$1933.52
$$
$1933.52
$$
Exercise 12
Step 1
1 of 4
[begin{gathered}
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
Step 2
2 of 4
Step 3
3 of 4
[begin{gathered}
{text{Notice that the total number of horizontal sticks }}{h_n}{text{ can be modelled}} hfill \
{text{by an arithmetic series with }}{t_1} = 2,{text{ and }}d = 1 hfill \
hfill \
{text{Using the formula for arithmetic series}} hfill \
{h_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {n + 3} right) = frac{1}{2}left( {{n^2} + 3n} right) hfill \
hfill \
{text{The total number of vertical sticks }}{v_n}{text{;has }} hfill \
{t_1} = 1{text{ and }}d = 1 hfill \
{;v_n} = frac{n}{2}left[ {2left( 2 right) + left( {n – 1} right)left( 1 right)} right] = frac{n}{2}left( {n + 3} right) =frac{ {n^2} + 3n }{2}hfill \
hfill \
{text{Therefore, the total number of sticks is}} hfill \
{S_n} = frac{1}{2}left( {{n^2} + 3n} right) + frac{left( {{n^2} + 3n} right)}{2} hfill \
{S_n} = n^2+3n hfill \
end{gathered} ]
\
The total number of sticks as a function of stack number is $S_n=n^2+3n$\\
{text{Notice that the total number of horizontal sticks }}{h_n}{text{ can be modelled}} hfill \
{text{by an arithmetic series with }}{t_1} = 2,{text{ and }}d = 1 hfill \
hfill \
{text{Using the formula for arithmetic series}} hfill \
{h_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {n + 3} right) = frac{1}{2}left( {{n^2} + 3n} right) hfill \
hfill \
{text{The total number of vertical sticks }}{v_n}{text{;has }} hfill \
{t_1} = 1{text{ and }}d = 1 hfill \
{;v_n} = frac{n}{2}left[ {2left( 2 right) + left( {n – 1} right)left( 1 right)} right] = frac{n}{2}left( {n + 3} right) =frac{ {n^2} + 3n }{2}hfill \
hfill \
{text{Therefore, the total number of sticks is}} hfill \
{S_n} = frac{1}{2}left( {{n^2} + 3n} right) + frac{left( {{n^2} + 3n} right)}{2} hfill \
{S_n} = n^2+3n hfill \
end{gathered} ]
\
The total number of sticks as a function of stack number is $S_n=n^2+3n$\\
Result
4 of 4
$$
S_n=n^2+3n
$$
S_n=n^2+3n
$$
Exercise 13
Step 1
1 of 3
[begin{gathered}
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
{mathbf{Identifying}},,{mathbf{Sequence}}:,,{text{For a sequence containing terms}} hfill \
{t_1},,{t_2},,{t_3},,{t_4},… hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_2} – {t_1} = {t_3} – {t_2} = {t_4} – {t_3} = d, hfill \
hfill \
{text{geometric }} Rightarrow {text{ }},frac{{{t_2}}}{{{t_1}}} = frac{{{t_3}}}{{{t_2}}} = frac{{{t_4}}}{{{t_3}}} = r hfill \
hfill \
{mathbf{General}},{mathbf{Term}},:,, hfill \
{text{arithmetic}} Rightarrow {text{ }},{t_n} = {t_1} + left( {n – 1} right)d hfill \
{text{geometric }} Rightarrow {text{ }}{t_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{mathbf{Recursive}},{mathbf{Formula}}: hfill \
hfill \
{text{arithmetic }} Rightarrow {text{ }}{t_1},,,,{t_n} = {t_{n – 1}} + d,,{text{where }}n > 1 hfill \
{text{geometric }} Rightarrow ,,,,{t_{1,}},,,,,,{t_n} = ,r cdot {t_{n – 1}},;{text{where }}n > 1, hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{Let}},{text{the numerator be }}{N_n}{text{ and denominator }}{D_n} hfill \
{N_n};{text{ is an arithmetic sequence with }}{t_1} = 1,,{text{and }}d = 1. hfill \
{text{Thus,}} hfill \
{N_n} = 1 + left( {n – 1} right)left( 1 right) = n hfill \
hfill \
{text{The denominator}},{text{is an arithmetic sequence with }} hfill \
{t_1} = 2,,{text{and}};d = 5 – 2 = 3 hfill \
{text{Thus,}} hfill \
{D_n} = 2 + left( {n – 1} right)left( 3 right) = 3n – 1 hfill \
hfill \
{text{Therefore, the general term for the original sequence is}} hfill \
{t_n} = frac{{{N_n}}}{{{D_n}}} = frac{n}{{3n – 1}} hfill \
hfill \
{text{With this general term, we can obtain the 100th term as}} hfill \
{t_{100}} = frac{{100}}{{3left( {100} right) – 1}} = frac{{100}}{{299}} hfill \
end{gathered} ]
{text{Let}},{text{the numerator be }}{N_n}{text{ and denominator }}{D_n} hfill \
{N_n};{text{ is an arithmetic sequence with }}{t_1} = 1,,{text{and }}d = 1. hfill \
{text{Thus,}} hfill \
{N_n} = 1 + left( {n – 1} right)left( 1 right) = n hfill \
hfill \
{text{The denominator}},{text{is an arithmetic sequence with }} hfill \
{t_1} = 2,,{text{and}};d = 5 – 2 = 3 hfill \
{text{Thus,}} hfill \
{D_n} = 2 + left( {n – 1} right)left( 3 right) = 3n – 1 hfill \
hfill \
{text{Therefore, the general term for the original sequence is}} hfill \
{t_n} = frac{{{N_n}}}{{{D_n}}} = frac{n}{{3n – 1}} hfill \
hfill \
{text{With this general term, we can obtain the 100th term as}} hfill \
{t_{100}} = frac{{100}}{{3left( {100} right) – 1}} = frac{{100}}{{299}} hfill \
end{gathered} ]
Result
3 of 3
$t_n=dfrac{n}{3n-1}$ ; $t_{100}=dfrac{100}{299}$
Exercise 14
Step 1
1 of 8
[begin{gathered}
{text{For an arithmetic series with first term }}{t_1}{text{ and common difference }}d hfill \
{text{The sum of the first }}n{text{ terms is}} hfill \
hfill \
{S_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {{t_1} + {t_n}} right) hfill \
end{gathered} ]
{text{For an arithmetic series with first term }}{t_1}{text{ and common difference }}d hfill \
{text{The sum of the first }}n{text{ terms is}} hfill \
hfill \
{S_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {{t_1} + {t_n}} right) hfill \
end{gathered} ]
Step 2
2 of 8
[begin{gathered}
{text{a}}{text{.) This is an arithmetic series with }}{t_1} = 1{text{ and }}d = 9 – 1 = 8 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( 1 right) + left( {50 – 1} right)left( 8 right)} right] = 9850 hfill \
end{gathered} ]
{text{a}}{text{.) This is an arithmetic series with }}{t_1} = 1{text{ and }}d = 9 – 1 = 8 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( 1 right) + left( {50 – 1} right)left( 8 right)} right] = 9850 hfill \
end{gathered} ]
Step 3
3 of 8
[begin{gathered}
{text{b}}{text{.) This is an arithmetic series with }}{t_1} = 21{text{ and }}d = 17 – 21 = – 4 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( {21} right) + left( {50 – 1} right)left( { – 4} right)} right] = – 3850 hfill \
end{gathered} ]
{text{b}}{text{.) This is an arithmetic series with }}{t_1} = 21{text{ and }}d = 17 – 21 = – 4 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( {21} right) + left( {50 – 1} right)left( { – 4} right)} right] = – 3850 hfill \
end{gathered} ]
Step 4
4 of 8
[begin{gathered}
{text{c}}{text{.) This is an arithmetic series with }}{t_1} = 31{text{ and }}d = 52 – 31 = 21 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( {31} right) + left( {50 – 1} right)left( {21} right)} right] = 27,275 hfill \
end{gathered} ]
{text{c}}{text{.) This is an arithmetic series with }}{t_1} = 31{text{ and }}d = 52 – 31 = 21 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( {31} right) + left( {50 – 1} right)left( {21} right)} right] = 27,275 hfill \
end{gathered} ]
Step 5
5 of 8
[begin{gathered}
{text{d}}{text{.) This is an arithmetic series with }}{t_1} = – 9{text{ and }}d = – 14 – left( { – 9} right) = – 5 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( { – 9} right) + left( {50 – 1} right)left( { – 5} right)} right] = – 6575 hfill \
end{gathered} ]
{text{d}}{text{.) This is an arithmetic series with }}{t_1} = – 9{text{ and }}d = – 14 – left( { – 9} right) = – 5 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( { – 9} right) + left( {50 – 1} right)left( { – 5} right)} right] = – 6575 hfill \
end{gathered} ]
Step 6
6 of 8
[begin{gathered}
{text{e}}{text{.) This is an arithmetic series with }}{t_1} = 17.5{text{ and }}d = 18.9 – 17.5 = 1.4 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( {17.5} right) + left( {50 – 1} right)left( {1.4} right)} right] = 2590 hfill \
end{gathered} ]
{text{e}}{text{.) This is an arithmetic series with }}{t_1} = 17.5{text{ and }}d = 18.9 – 17.5 = 1.4 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( {17.5} right) + left( {50 – 1} right)left( {1.4} right)} right] = 2590 hfill \
end{gathered} ]
Step 7
7 of 8
[begin{gathered}
{text{f}}{text{.) This is an arithmetic series with }}{t_1} = – 39{text{ and }}d = – 31 – left( { – 39} right) = 8 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( { – 39} right) + left( {50 – 1} right)left( 8 right)} right] = 7850 hfill \
end{gathered} ]
{text{f}}{text{.) This is an arithmetic series with }}{t_1} = – 39{text{ and }}d = – 31 – left( { – 39} right) = 8 hfill \
{S_{50}} = frac{{50}}{2}left[ {2left( { – 39} right) + left( {50 – 1} right)left( 8 right)} right] = 7850 hfill \
end{gathered} ]
Result
8 of 8
a.) 9850
b.) $-$3850
c.) 27 275
d.) $-6$575
e.) 2590
f.) 7850
Exercise 15
Step 1
1 of 8
[begin{gathered}
{text{For an arithmetic series with first term }}{t_1}{text{ and common difference }}d hfill \
{text{The sum of the first }}n{text{ terms is}} hfill \
hfill \
{S_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {{t_1} + {t_n}} right) hfill \
end{gathered} ]\\
If either $d$ or $t_1$ is not given, you can use the fact that if $t_j$ and $t_k$ are terms of an arithmetic sequence, then\\
$t_k-t_j=(k-j)d$
{text{For an arithmetic series with first term }}{t_1}{text{ and common difference }}d hfill \
{text{The sum of the first }}n{text{ terms is}} hfill \
hfill \
{S_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {{t_1} + {t_n}} right) hfill \
end{gathered} ]\\
If either $d$ or $t_1$ is not given, you can use the fact that if $t_j$ and $t_k$ are terms of an arithmetic sequence, then\\
$t_k-t_j=(k-j)d$
Step 2
2 of 8
[begin{gathered}
{text{a}}{text{.)}},{text{This is an arithmetic series with }} hfill \
{t_1} = 24 hfill \
d = 11 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {24} right) + left( {25 – 1} right)left( {11} right)} right] = 3900 hfill \
end{gathered} ]
{text{a}}{text{.)}},{text{This is an arithmetic series with }} hfill \
{t_1} = 24 hfill \
d = 11 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {24} right) + left( {25 – 1} right)left( {11} right)} right] = 3900 hfill \
end{gathered} ]
Step 3
3 of 8
[begin{gathered}
{text{b}}{text{.) This is an arithmetic series with}} hfill \
{t_{25}} = 374 hfill \
{t_1} = 91 hfill \
{S_{25}} = frac{{25}}{2}left( {91 + 374} right) = 5812.5 hfill \
end{gathered} ]
{text{b}}{text{.) This is an arithmetic series with}} hfill \
{t_{25}} = 374 hfill \
{t_1} = 91 hfill \
{S_{25}} = frac{{25}}{2}left( {91 + 374} right) = 5812.5 hfill \
end{gathered} ]
Step 4
4 of 8
[begin{gathered}
{text{c}}{text{.) This is an arithmetic series with}} hfill \
{t_1} = 84 hfill \
d = 57 – 84 = – 27 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {84} right) + left( {25 – 1} right)left( { – 27} right)} right] = – 6000 hfill \
end{gathered} ]
{text{c}}{text{.) This is an arithmetic series with}} hfill \
{t_1} = 84 hfill \
d = 57 – 84 = – 27 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {84} right) + left( {25 – 1} right)left( { – 27} right)} right] = – 6000 hfill \
end{gathered} ]
Step 5
5 of 8
[begin{gathered}
{text{d}}{text{.) This is an arithmetic series with}} hfill \
{t_3} = 43 hfill \
d = – 11 hfill \
{text{To solve for the first term,}} hfill \
{t_3} – {t_1} = left( {3 – 1} right)d,, Rightarrow ,,{t_1} = 43 – 2left( { – 11} right) = 64 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {64} right) + left( {25 – 1} right)left( { – 11} right)} right] = – 1700 hfill \
end{gathered} ]
{text{d}}{text{.) This is an arithmetic series with}} hfill \
{t_3} = 43 hfill \
d = – 11 hfill \
{text{To solve for the first term,}} hfill \
{t_3} – {t_1} = left( {3 – 1} right)d,, Rightarrow ,,{t_1} = 43 – 2left( { – 11} right) = 64 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {64} right) + left( {25 – 1} right)left( { – 11} right)} right] = – 1700 hfill \
end{gathered} ]
Step 6
6 of 8
[begin{gathered}
{text{e}}{text{.)}},,{text{This is an arithmetic series with}} hfill \
d = – 4 hfill \
{t_{12}} = 19 hfill \
{text{To solve for the first term,}} hfill \
{t_{12}} – {t_1} = left( {12 – 1} right)d Rightarrow {t_1} = 19 – 11left( { – 4} right) = 63 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {63} right) + left( {25 – 1} right)left( { – 4} right)} right] = 375 hfill \
end{gathered} ]
{text{e}}{text{.)}},,{text{This is an arithmetic series with}} hfill \
d = – 4 hfill \
{t_{12}} = 19 hfill \
{text{To solve for the first term,}} hfill \
{t_{12}} – {t_1} = left( {12 – 1} right)d Rightarrow {t_1} = 19 – 11left( { – 4} right) = 63 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {63} right) + left( {25 – 1} right)left( { – 4} right)} right] = 375 hfill \
end{gathered} ]
Step 7
7 of 8
[begin{gathered}
{text{f}}{text{.)}},,{text{This is an arithmetic series with}} hfill \
{t_5} = 142 hfill \
{t_{15}} = 12 hfill \
{text{To solve for the common difference}} hfill \
{t_{15}} – {t_5} = left( {15 – 5} right)d Rightarrow 142 – 12 = 10d Rightarrow ,d = – 13 hfill \
hfill \
{text{To solve for the first term}} hfill \
{t_5} – {t_1} = left( {5 – 1} right)left( d right) Rightarrow {t_1} = 142 – 4left( { – 13} right) = 194 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {194} right) + left( {25 – 1} right)left( { – 13} right)} right] = 950 hfill \
end{gathered} ]
{text{f}}{text{.)}},,{text{This is an arithmetic series with}} hfill \
{t_5} = 142 hfill \
{t_{15}} = 12 hfill \
{text{To solve for the common difference}} hfill \
{t_{15}} – {t_5} = left( {15 – 5} right)d Rightarrow 142 – 12 = 10d Rightarrow ,d = – 13 hfill \
hfill \
{text{To solve for the first term}} hfill \
{t_5} – {t_1} = left( {5 – 1} right)left( d right) Rightarrow {t_1} = 142 – 4left( { – 13} right) = 194 hfill \
{S_{25}} = frac{{25}}{2}left[ {2left( {194} right) + left( {25 – 1} right)left( { – 13} right)} right] = 950 hfill \
end{gathered} ]
Result
8 of 8
a.) 3900
b.) 5812.5
c.) $-$6000
d.) $-$1700
e.) 375
f.) 950
Exercise 16
Step 1
1 of 5
[begin{gathered}
{text{For an arithmetic series with first term }}{t_1}{text{ and common difference }}d hfill \
{text{The sum of the first }}n{text{ terms is}} hfill \
hfill \
{S_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {{t_1} + {t_n}} right) hfill \
end{gathered} ]\\
{text{For an arithmetic series with first term }}{t_1}{text{ and common difference }}d hfill \
{text{The sum of the first }}n{text{ terms is}} hfill \
hfill \
{S_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {{t_1} + {t_n}} right) hfill \
end{gathered} ]\\
Step 2
2 of 5
[begin{gathered}
{text{a}}{text{.)}},{text{This is an arithmetic series with }} hfill \
{t_1} = 1 hfill \
d = 13 – 1 = 12 hfill \\
{text{General Term: }}{t_n} = 1 + left( {n – 1} right)left( {12} right) = 12n – 11 hfill \\
{text{We shall find }}n{text{ such that }}{t_n} = 145 hfill \\
145 = 12n – 11 Rightarrow n = frac{{145 + 11}}{{12}} = 13 hfill \\
{text{Thus, the sum of the series is }}{S_{13}} hfill \\
{S_{13}} = frac{{13}}{2}left( {1 + 145} right) = 949 hfill \\
end{gathered} ]
{text{a}}{text{.)}},{text{This is an arithmetic series with }} hfill \
{t_1} = 1 hfill \
d = 13 – 1 = 12 hfill \\
{text{General Term: }}{t_n} = 1 + left( {n – 1} right)left( {12} right) = 12n – 11 hfill \\
{text{We shall find }}n{text{ such that }}{t_n} = 145 hfill \\
145 = 12n – 11 Rightarrow n = frac{{145 + 11}}{{12}} = 13 hfill \\
{text{Thus, the sum of the series is }}{S_{13}} hfill \\
{S_{13}} = frac{{13}}{2}left( {1 + 145} right) = 949 hfill \\
end{gathered} ]
Step 3
3 of 5
[begin{gathered}
{text{b}}{text{.)}},{text{This is an arithmetic series with }} hfill \
{t_1} = 9 hfill \
d = 42 – 9 = 33 hfill \\
{text{General Term: }}{t_n} = 9 + left( {n – 1} right)left( {33} right) = 33n – 24 hfill \\
{text{We shall find }}n{text{ such that }}{t_n} = 4068 hfill \\
4068 = 33n – 24 Rightarrow n = frac{{4068 + 24}}{{33}} = 124 hfill \\
{text{Thus, the sum of the series is }}{S_{124}} hfill \\
{S_{124}} = frac{{124}}{2}left( {9 + 4068} right) = 252,774 hfill \\
end{gathered} ]
{text{b}}{text{.)}},{text{This is an arithmetic series with }} hfill \
{t_1} = 9 hfill \
d = 42 – 9 = 33 hfill \\
{text{General Term: }}{t_n} = 9 + left( {n – 1} right)left( {33} right) = 33n – 24 hfill \\
{text{We shall find }}n{text{ such that }}{t_n} = 4068 hfill \\
4068 = 33n – 24 Rightarrow n = frac{{4068 + 24}}{{33}} = 124 hfill \\
{text{Thus, the sum of the series is }}{S_{124}} hfill \\
{S_{124}} = frac{{124}}{2}left( {9 + 4068} right) = 252,774 hfill \\
end{gathered} ]
Step 4
4 of 5
[begin{gathered}
{text{c}}{text{.)}},{text{This is an arithmetic series with }} hfill \
{t_1} = 123 hfill \
d = 118 – 123 = – 5 hfill \\
{text{General Term: }}{t_n} = 123 + left( {n – 1} right)left( { – 5} right) = 128 – 5n hfill \\
{text{We shall find }}n{text{ such that }}{t_n} = – 122 hfill \\
– 122 = 128 – 5n Rightarrow n = frac{{128 + 122}}{5} = 50 hfill \\
{text{Thus, the sum of the series is }}{S_{50}} hfill \\
{S_{50}} = frac{{50}}{2}left( {123 – 122} right) = 25 hfill \ \
end{gathered} ]
{text{c}}{text{.)}},{text{This is an arithmetic series with }} hfill \
{t_1} = 123 hfill \
d = 118 – 123 = – 5 hfill \\
{text{General Term: }}{t_n} = 123 + left( {n – 1} right)left( { – 5} right) = 128 – 5n hfill \\
{text{We shall find }}n{text{ such that }}{t_n} = – 122 hfill \\
– 122 = 128 – 5n Rightarrow n = frac{{128 + 122}}{5} = 50 hfill \\
{text{Thus, the sum of the series is }}{S_{50}} hfill \\
{S_{50}} = frac{{50}}{2}left( {123 – 122} right) = 25 hfill \ \
end{gathered} ]
Result
5 of 5
a.) 949
b.) 252 774
c.) 25
Exercise 17
Step 1
1 of 3
[begin{gathered}
{text{For an arithmetic series with first term }}{t_1}{text{ and common difference }}d hfill \
{text{The sum of the first }}n{text{ terms is}} hfill \
hfill \
{S_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {{t_1} + {t_n}} right) hfill \
end{gathered} ]\\
{text{For an arithmetic series with first term }}{t_1}{text{ and common difference }}d hfill \
{text{The sum of the first }}n{text{ terms is}} hfill \
hfill \
{S_n} = frac{n}{2}left[ {2{t_1} + left( {n – 1} right)d} right] = frac{n}{2}left( {{t_1} + {t_n}} right) hfill \
end{gathered} ]\\
Step 2
2 of 3
The distance that the space craft descended decreases by $7$ m each second starting from $64$ m in the first second. Thus, the total distance descended can be modeled by an arithmetic series with $t_1=64$ and $d=-7$.
On the 10th second,
$S_{10}=dfrac{10}{2}[2(64)+(10-1)(-7)]=325$
Therefore, on the 10th second, the total distance at which the spacecraft has descended is $boxed{bold{325;meters}}$
Result
3 of 3
$325$ m
Exercise 18
Step 1
1 of 8
[begin{gathered}
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
Step 2
2 of 8
[begin{gathered}
{text{a}}{text{.) This is a geometric series with}} hfill \
r = frac{{33}}{{11}} = 3 hfill \
{t_1} = 11 hfill \
{text{General term: }}{t_n} = 11 cdot {left( 3 right)^{n – 1}} hfill \
{text{Sixth term: }}{{text{t}}_6} = 11 cdot {3^{6 – 1}} = 2673 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 11 cdot frac{{{3^6} – 1}}{{3 – 1}} = 4004 hfill \
end{gathered} ]
{text{a}}{text{.) This is a geometric series with}} hfill \
r = frac{{33}}{{11}} = 3 hfill \
{t_1} = 11 hfill \
{text{General term: }}{t_n} = 11 cdot {left( 3 right)^{n – 1}} hfill \
{text{Sixth term: }}{{text{t}}_6} = 11 cdot {3^{6 – 1}} = 2673 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 11 cdot frac{{{3^6} – 1}}{{3 – 1}} = 4004 hfill \
end{gathered} ]
Step 3
3 of 8
[begin{gathered}
{text{b}}{text{.) This is a geometric series with}} hfill \
r = frac{{1.11111}}{{0.111,111}} = 10 hfill \
{t_1} = 0.111,111 hfill \
{text{General term: }}{t_n} = 0.111111 cdot {10^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = 0.111,111{left( {10} right)^{6 – 1}} = 11,111.1 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 0.111,111frac{{{{10}^6} – 1}}{{10 – 1}} approx 12,345.6543 hfill \
end{gathered} ]
{text{b}}{text{.) This is a geometric series with}} hfill \
r = frac{{1.11111}}{{0.111,111}} = 10 hfill \
{t_1} = 0.111,111 hfill \
{text{General term: }}{t_n} = 0.111111 cdot {10^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = 0.111,111{left( {10} right)^{6 – 1}} = 11,111.1 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 0.111,111frac{{{{10}^6} – 1}}{{10 – 1}} approx 12,345.6543 hfill \
end{gathered} ]
Step 4
4 of 8
[begin{gathered}
{text{c}}{text{.) This is a geometric series with}} hfill \
r = frac{{ – 12}}{6} = – 2 hfill \
{t_1} = 6 hfill \
{text{General term: }}{t_n} = 6 cdot {left( { – 2} right)^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = 6{left( { – 2} right)^{6 – 1}} = – 192 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 6frac{{{{left( { – 2} right)}^6} – 1}}{{left( { – 2} right) – 1}} = – 126 hfill \
end{gathered} ]
{text{c}}{text{.) This is a geometric series with}} hfill \
r = frac{{ – 12}}{6} = – 2 hfill \
{t_1} = 6 hfill \
{text{General term: }}{t_n} = 6 cdot {left( { – 2} right)^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = 6{left( { – 2} right)^{6 – 1}} = – 192 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 6frac{{{{left( { – 2} right)}^6} – 1}}{{left( { – 2} right) – 1}} = – 126 hfill \
end{gathered} ]
Step 5
5 of 8
[begin{gathered}
{text{d}}{text{.) This is a geometric series with}} hfill \
r = frac{{21,870}}{{32,805}} = frac{2}{3} hfill \
{t_1} = 32,805 hfill \
{text{General term: }}{t_n} = 32,805 cdot {left( {frac{2}{3}} right)^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = 32,805 cdot {left( {frac{2}{3}} right)^{6 – 1}} = 4320 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 32,805frac{{{{left( {2/3} right)}^6} – 1}}{{left( {2/3} right) – 1}} = 89,775 hfill \
end{gathered} ]
{text{d}}{text{.) This is a geometric series with}} hfill \
r = frac{{21,870}}{{32,805}} = frac{2}{3} hfill \
{t_1} = 32,805 hfill \
{text{General term: }}{t_n} = 32,805 cdot {left( {frac{2}{3}} right)^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = 32,805 cdot {left( {frac{2}{3}} right)^{6 – 1}} = 4320 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 32,805frac{{{{left( {2/3} right)}^6} – 1}}{{left( {2/3} right) – 1}} = 89,775 hfill \
end{gathered} ]
Step 6
6 of 8
[begin{gathered}
{text{e}}{text{.) This is a geometric series with}} hfill \
r = frac{{ – 25.5}}{{17}} = – frac{3}{2} hfill \
{t_1} = 17 hfill \
{text{General term: }}{t_n} = 17 cdot {left( { – frac{3}{2}} right)^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = 17 cdot {left( { – frac{3}{2}} right)^{6 – 1}} = – 129.09375 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 17 cdot frac{{{{left( { – 3/2} right)}^6} – 1}}{{left( { – 3/2} right) – 1}} = – 70.65625 hfill \
end{gathered} ]
{text{e}}{text{.) This is a geometric series with}} hfill \
r = frac{{ – 25.5}}{{17}} = – frac{3}{2} hfill \
{t_1} = 17 hfill \
{text{General term: }}{t_n} = 17 cdot {left( { – frac{3}{2}} right)^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = 17 cdot {left( { – frac{3}{2}} right)^{6 – 1}} = – 129.09375 hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = 17 cdot frac{{{{left( { – 3/2} right)}^6} – 1}}{{left( { – 3/2} right) – 1}} = – 70.65625 hfill \
end{gathered} ]
Step 7
7 of 8
[begin{gathered}
{text{f}}{text{.) This is a geometric series with}} hfill \
r = frac{{3/10}}{{1/2}} = frac{3}{5} hfill \
{t_1} = frac{1}{2} hfill \
{text{General term: }}{t_n} = frac{1}{2} cdot {left( {frac{3}{5}} right)^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = frac{1}{2} cdot {left( {frac{3}{5}} right)^{6 – 1}} = frac{{243}}{{6250}} hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = frac{1}{2} cdot frac{{{{left( {3/5} right)}^6} – 1}}{{left( {3/5} right) – 1}} = frac{{3724}}{{3125}} hfill \
end{gathered} ]
{text{f}}{text{.) This is a geometric series with}} hfill \
r = frac{{3/10}}{{1/2}} = frac{3}{5} hfill \
{t_1} = frac{1}{2} hfill \
{text{General term: }}{t_n} = frac{1}{2} cdot {left( {frac{3}{5}} right)^{n – 1}} hfill \
{text{Sixth term: }}{t_6} = frac{1}{2} cdot {left( {frac{3}{5}} right)^{6 – 1}} = frac{{243}}{{6250}} hfill \
{text{Sum of 6 terms: }}{{text{S}}_6} = frac{1}{2} cdot frac{{{{left( {3/5} right)}^6} – 1}}{{left( {3/5} right) – 1}} = frac{{3724}}{{3125}} hfill \
end{gathered} ]
Result
8 of 8
a.) $4004$
b.) $12345.6543$
c.) $-126$
d.) $89;775$
e.) $-70.65625$
f.) $dfrac{3724}{3125}$
Exercise 19
Step 1
1 of 6
[begin{gathered}
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
Step 2
2 of 6
[begin{gathered}
{text{a}}{text{.) This is a geometric series with}} hfill \
r = 4 hfill \
{t_1} = – 6 hfill \
{text{The sum of the first 8 terms is}} hfill \
{S_8} = left( { – 6} right) cdot frac{{{4^8} – 1}}{{4 – 1}} = – 131,,070 hfill \
end{gathered} ]
{text{a}}{text{.) This is a geometric series with}} hfill \
r = 4 hfill \
{t_1} = – 6 hfill \
{text{The sum of the first 8 terms is}} hfill \
{S_8} = left( { – 6} right) cdot frac{{{4^8} – 1}}{{4 – 1}} = – 131,,070 hfill \
end{gathered} ]
Step 3
3 of 6
[begin{gathered}
{text{b}}{text{.) This is a geometric series with}} hfill \
{t_1} = 42 hfill \
{t_9} = 2112 hfill \
{text{We shall find }}r hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
2112 = 42 cdot {r^{9 – 1}} Rightarrow r = {left( {frac{{2112}}{{42}}} right)^{1/8}} = 1.631851286 hfill \
{text{The sum of the first 8 terms is}} hfill \
{S_8} = 42 cdot frac{{{{left( {1.631851286} right)}^8} – 1}}{{1.631851286 – 1}} = 3276.087344 hfill \
end{gathered} ]
{text{b}}{text{.) This is a geometric series with}} hfill \
{t_1} = 42 hfill \
{t_9} = 2112 hfill \
{text{We shall find }}r hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
2112 = 42 cdot {r^{9 – 1}} Rightarrow r = {left( {frac{{2112}}{{42}}} right)^{1/8}} = 1.631851286 hfill \
{text{The sum of the first 8 terms is}} hfill \
{S_8} = 42 cdot frac{{{{left( {1.631851286} right)}^8} – 1}}{{1.631851286 – 1}} = 3276.087344 hfill \
end{gathered} ]
Step 4
4 of 6
[begin{gathered}
{text{c}}{text{.) This is a geometric series with}} hfill \
r = frac{{{t_2}}}{{{t_1}}} = frac{{80}}{{320}} = frac{1}{4} hfill \
{t_1} = 320 hfill \
{text{The sum of the first 8 terms is}} hfill \
{S_8} = 320 cdot frac{{{{left( {1/4} right)}^8} – 1}}{{left( {1/4} right) – 1}} = 426.6601563 hfill \
end{gathered} ]
{text{c}}{text{.) This is a geometric series with}} hfill \
r = frac{{{t_2}}}{{{t_1}}} = frac{{80}}{{320}} = frac{1}{4} hfill \
{t_1} = 320 hfill \
{text{The sum of the first 8 terms is}} hfill \
{S_8} = 320 cdot frac{{{{left( {1/4} right)}^8} – 1}}{{left( {1/4} right) – 1}} = 426.6601563 hfill \
end{gathered} ]
Step 5
5 of 6
[begin{gathered}
{text{d}}{text{.) This is a geometric series with}} hfill \
r = 5 hfill \
{t_3} = 35 hfill \
{text{We shall find }}{t_1} hfill \
{t_3} = {t_1} cdot {r^{n – 1}} Rightarrow {t_1} = frac{{{t_3}}}{{{r^{n – 1}}}} = frac{{35}}{{{5^{3 – 1}}}} = frac{7}{5} hfill \
{text{The sum of the first 8 terms is}} hfill \
{S_8} = frac{7}{5} cdot frac{{{{left( 5 right)}^8} – 1}}{{left( 5 right) – 1}} = 136718.4 hfill \
end{gathered} ]
{text{d}}{text{.) This is a geometric series with}} hfill \
r = 5 hfill \
{t_3} = 35 hfill \
{text{We shall find }}{t_1} hfill \
{t_3} = {t_1} cdot {r^{n – 1}} Rightarrow {t_1} = frac{{{t_3}}}{{{r^{n – 1}}}} = frac{{35}}{{{5^{3 – 1}}}} = frac{7}{5} hfill \
{text{The sum of the first 8 terms is}} hfill \
{S_8} = frac{7}{5} cdot frac{{{{left( 5 right)}^8} – 1}}{{left( 5 right) – 1}} = 136718.4 hfill \
end{gathered} ]
Result
6 of 6
a.) $-131;070$
b.) $3276.087344$
c.) 426.6601563
d.) 136718.4
Exercise 20
Step 1
1 of 3
[begin{gathered}
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
Step 2
2 of 3
Observe that the number of filled orders goes like
$15+30+60+…$
This is a geometric series with $t_1=15$ and $r=2$.
Since 1 year has 12 months, we shall find $S_{12}$
$S_{12}=15cdot dfrac{2^{12}-1}{2-1}=61;425$
Therefore, a total of $boxed{bold{61,425 ;orders}}$ are filled in the first year.
Result
3 of 3
$61,425$ orders
Exercise 21
Step 1
1 of 3
[begin{gathered}
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{Condition: }}, hfill \
{text{In the arithmetic sequence }}{t_1},,{t_2},;{t_3},… hfill \
{text{The terms }}{t_1},,{t_5},,{text{ and }}{t_{13}};{text{are the first three terms of}} hfill \
{text{a geometric sequence with }}r = 2. hfill \
{text{We shall find the general term for this sequence}}{text{.}} hfill \
hfill \
{text{From the property of geometric sequence}} hfill \
left{ {begin{array}{*{20}{c}}
{{t_5} = 2{t_1}} \
{{t_{13}} = 2{t_5}}
end{array}} right. hfill \
hfill \
{text{We will then express }}{t_5}{text{ using arithmetic sequence}} hfill \
{t_5} = 2{t_1} hfill \
{t_n} = {t_1} + left( {n – 1} right)d hfill \
{t_5} = {t_1} + left( {5 – 1} right)d hfill \
2{t_1} = {t_1} + left( {5 – 1} right)d hfill \
{t_1} = 4d hfill \
hfill \
{text{Knowing that }}{t_{21}} = 72,{text{ we can solve for }},d hfill \
{t_{21}} = {t_1} + left( {21 – 1} right)d hfill \
72 = 4d + 20d hfill \
d = frac{{72}}{{24}} = 3 hfill \
hfill \
{text{Since }}{t_1} = 4d,,, hfill \
Rightarrow ,,{t_1} = 4left( 3 right) = 12 hfill \
hfill \
{text{Since we know }}{t_1}{text{ and }}r,,{text{we can now solve for }} hfill \
{S_{10}}{text{ of the geometric series}} hfill \
{S_{10}} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
{S_{10}} = 12 cdot frac{{{2^{10}} – 1}}{{2 – 1}} = boxed{12,276} hfill \
end{gathered} ]
{text{Condition: }}, hfill \
{text{In the arithmetic sequence }}{t_1},,{t_2},;{t_3},… hfill \
{text{The terms }}{t_1},,{t_5},,{text{ and }}{t_{13}};{text{are the first three terms of}} hfill \
{text{a geometric sequence with }}r = 2. hfill \
{text{We shall find the general term for this sequence}}{text{.}} hfill \
hfill \
{text{From the property of geometric sequence}} hfill \
left{ {begin{array}{*{20}{c}}
{{t_5} = 2{t_1}} \
{{t_{13}} = 2{t_5}}
end{array}} right. hfill \
hfill \
{text{We will then express }}{t_5}{text{ using arithmetic sequence}} hfill \
{t_5} = 2{t_1} hfill \
{t_n} = {t_1} + left( {n – 1} right)d hfill \
{t_5} = {t_1} + left( {5 – 1} right)d hfill \
2{t_1} = {t_1} + left( {5 – 1} right)d hfill \
{t_1} = 4d hfill \
hfill \
{text{Knowing that }}{t_{21}} = 72,{text{ we can solve for }},d hfill \
{t_{21}} = {t_1} + left( {21 – 1} right)d hfill \
72 = 4d + 20d hfill \
d = frac{{72}}{{24}} = 3 hfill \
hfill \
{text{Since }}{t_1} = 4d,,, hfill \
Rightarrow ,,{t_1} = 4left( 3 right) = 12 hfill \
hfill \
{text{Since we know }}{t_1}{text{ and }}r,,{text{we can now solve for }} hfill \
{S_{10}}{text{ of the geometric series}} hfill \
{S_{10}} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
{S_{10}} = 12 cdot frac{{{2^{10}} – 1}}{{2 – 1}} = boxed{12,276} hfill \
end{gathered} ]
Result
3 of 3
$$
S_{10}=12,276
$$
S_{10}=12,276
$$
Exercise 22
Step 1
1 of 7
[begin{gathered}
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
{text{The sum of the first }}n{text{ terms in geometric series with}} hfill \
{text{first term }}{t_1}{text{ and common ratio }}r{text{ is}} hfill \
hfill \
{{text{S}}_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
hfill \
{text{If }}r{text{ or }}{t_1}{text{ is not given, you can solve them using}} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
end{gathered} ]
Step 2
2 of 7
[begin{gathered}
{text{a}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = 7 hfill \
r = frac{{14}}{7} = 2 hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = 7 cdot {2^{n – 1}} hfill \
hfill \
{text{We shall find }}n{text{ such that }}{t_n} = 3584 hfill \
{t_n} = 7 cdot {2^{n – 1}} hfill \
3584 = 7 cdot {2^{n – 1}} hfill \
512 = {2^{n – 1}} hfill \
{2^9} = {2^{n – 1}} hfill \
n = 10 hfill \
hfill \
{text{The sum of 10 terms is}} hfill \
{S_{10}} = 7 cdot frac{{{2^{10}} – 1}}{{2 – 1}} = 7161 hfill \
end{gathered} ]
{text{a}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = 7 hfill \
r = frac{{14}}{7} = 2 hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = 7 cdot {2^{n – 1}} hfill \
hfill \
{text{We shall find }}n{text{ such that }}{t_n} = 3584 hfill \
{t_n} = 7 cdot {2^{n – 1}} hfill \
3584 = 7 cdot {2^{n – 1}} hfill \
512 = {2^{n – 1}} hfill \
{2^9} = {2^{n – 1}} hfill \
n = 10 hfill \
hfill \
{text{The sum of 10 terms is}} hfill \
{S_{10}} = 7 cdot frac{{{2^{10}} – 1}}{{2 – 1}} = 7161 hfill \
end{gathered} ]
Step 3
3 of 7
[begin{gathered}
{text{b}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = – 3 hfill \
r = frac{{ – 6}}{{ – 3}} = 2 hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = left( { – 3} right) cdot {2^{n – 1}} hfill \
hfill \
{text{We shall find }}n{text{ such that }}{t_n} = – 768 hfill \
{t_n} = left( { – 3} right) cdot {2^{n – 1}} hfill \
– 768 = left( { – 3} right) cdot {2^{n – 1}} hfill \
256 = {2^{n – 1}} hfill \
{2^8} = {2^{n – 1}} hfill \
n = 9 hfill \
hfill \
{text{The sum of 9 terms is}} hfill \
{S_9} = left( { – 3} right) cdot frac{{{2^9} – 1}}{{2 – 1}} = – 1533 hfill \
end{gathered} ]
{text{b}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = – 3 hfill \
r = frac{{ – 6}}{{ – 3}} = 2 hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = left( { – 3} right) cdot {2^{n – 1}} hfill \
hfill \
{text{We shall find }}n{text{ such that }}{t_n} = – 768 hfill \
{t_n} = left( { – 3} right) cdot {2^{n – 1}} hfill \
– 768 = left( { – 3} right) cdot {2^{n – 1}} hfill \
256 = {2^{n – 1}} hfill \
{2^8} = {2^{n – 1}} hfill \
n = 9 hfill \
hfill \
{text{The sum of 9 terms is}} hfill \
{S_9} = left( { – 3} right) cdot frac{{{2^9} – 1}}{{2 – 1}} = – 1533 hfill \
end{gathered} ]
Step 4
4 of 7
[begin{gathered}
{text{c}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = 1 hfill \
r = frac{5}{2} hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = {left( {frac{5}{2}} right)^{n – 1}} hfill \
hfill \
{text{We shall find }}n{text{ such that }}{t_n} = frac{{15625}}{{64}} hfill \
{t_n} = {left( {frac{5}{2}} right)^{n – 1}} hfill \
frac{{15625}}{{64}} = {left( {frac{5}{2}} right)^{n – 1}} hfill \
frac{{{5^6}}}{{{2^6}}} = {left( {frac{5}{2}} right)^{n – 1}} hfill \
6 = n – 1 hfill \
n = 7 hfill \
hfill \
{text{The sum of 7 terms is}} hfill \
{S_9} = 1 cdot frac{{{{left( {5/2} right)}^7} – 1}}{{left( {5/2} right) – 1}} = frac{{25,999}}{{64}} hfill \
end{gathered} ]
{text{c}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = 1 hfill \
r = frac{5}{2} hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = {left( {frac{5}{2}} right)^{n – 1}} hfill \
hfill \
{text{We shall find }}n{text{ such that }}{t_n} = frac{{15625}}{{64}} hfill \
{t_n} = {left( {frac{5}{2}} right)^{n – 1}} hfill \
frac{{15625}}{{64}} = {left( {frac{5}{2}} right)^{n – 1}} hfill \
frac{{{5^6}}}{{{2^6}}} = {left( {frac{5}{2}} right)^{n – 1}} hfill \
6 = n – 1 hfill \
n = 7 hfill \
hfill \
{text{The sum of 7 terms is}} hfill \
{S_9} = 1 cdot frac{{{{left( {5/2} right)}^7} – 1}}{{left( {5/2} right) – 1}} = frac{{25,999}}{{64}} hfill \
end{gathered} ]
Step 5
5 of 7
[begin{gathered}
{text{d}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = 96000 hfill \
r = – frac{1}{2} hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = 9600{left( { – frac{1}{2}} right)^{n – 1}} hfill \
hfill \
{text{We shall find }}n{text{ such that }}{t_n} = 375 hfill \
375 = 96000{left( { – frac{1}{2}} right)^{n – 1}} hfill \
frac{1}{{256}} = {left( { – frac{1}{2}} right)^{n – 1}} hfill \
frac{1}{{{2^8}}} = {left( { – frac{1}{2}} right)^{n – 1}} hfill \
8 = n – 1 hfill \
n = 9 hfill \
hfill \
{text{The sum of 9 terms is}} hfill \
{S_9} = 96000frac{{{{left( { – 1/2} right)}^9} – 1}}{{left( { – 1/2} right) – 1}} = 64125 hfill \
end{gathered} ]
{text{d}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = 96000 hfill \
r = – frac{1}{2} hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = 9600{left( { – frac{1}{2}} right)^{n – 1}} hfill \
hfill \
{text{We shall find }}n{text{ such that }}{t_n} = 375 hfill \
375 = 96000{left( { – frac{1}{2}} right)^{n – 1}} hfill \
frac{1}{{256}} = {left( { – frac{1}{2}} right)^{n – 1}} hfill \
frac{1}{{{2^8}}} = {left( { – frac{1}{2}} right)^{n – 1}} hfill \
8 = n – 1 hfill \
n = 9 hfill \
hfill \
{text{The sum of 9 terms is}} hfill \
{S_9} = 96000frac{{{{left( { – 1/2} right)}^9} – 1}}{{left( { – 1/2} right) – 1}} = 64125 hfill \
end{gathered} ]
Step 6
6 of 7
[begin{gathered}
{text{d}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = 1000 hfill \
r = 1.06 hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = 1000{left( {1.06} right)^{n – 1}} hfill \
hfill \
{text{We shall find }}{S_{13}} hfill \
hfill \
{S_{13}} = 1000 cdot frac{{{{1.06}^{13}} – 1}}{{1.06 – 1}} = 18882.13767 hfill \
end{gathered} ]
{text{d}}{text{.) This is a geometric sequence with }} hfill \
{t_1} = 1000 hfill \
r = 1.06 hfill \
hfill \
{text{General term: }}{t_n} = {t_1} cdot {r^{n = 1}} = 1000{left( {1.06} right)^{n – 1}} hfill \
hfill \
{text{We shall find }}{S_{13}} hfill \
hfill \
{S_{13}} = 1000 cdot frac{{{{1.06}^{13}} – 1}}{{1.06 – 1}} = 18882.13767 hfill \
end{gathered} ]
Result
7 of 7
a.) 7161
b.) -1533
c.) $dfrac{25999}{64}$
d.) 64125
e.) 18882.14
Exercise 23
Step 1
1 of 9
$$
begin{array}{c}
{{color{#4257b2}text{row 0}}}&{}&{}&{}&{}&{}&{}&1&{}&{}&{}&{}&{}&{}&{} \
{{color{#4257b2}text{row 1}}}&{}&{}&{}&{}&{}&1&{}&1&{}&{}&{}&{}&{}&{} \
{{color{#4257b2}text{row 2}}}&{}&{}&{}&{}&1&{}&2&{}&1&{}&{}&{}&{}&{} \
{{color{#4257b2}text{row 3}}}&{}&{}&{}&1&{}&3&{}&3&{}&1&{}&{}&{}&{} \
{{color{#4257b2}text{row 4}}}&{}&{}&1&{}&4&{}&6&{}&4&{}&1&{}&{}&{} \
{{color{#4257b2}text{row 5}}}&{}&1&{}&5&{}&{10}&{}&{10}&{}&5&{}&1&{}&{} \
{{color{#4257b2}text{row 6}}}&1&{}&6&{}&{15}&{}&{20}&{}&{15}&{}&6&{}&1&{} \
{}&{}&{}&{}&{}&{}&{}&.&{}&{}&{}&{}&{}&{}&{} \
{}&{}&{}&{}&{}&{}&{}&.&{}&{}&{}&{}&{}&{}&{} \
{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{} \
end{array}
$$
begin{array}{c}
{{color{#4257b2}text{row 0}}}&{}&{}&{}&{}&{}&{}&1&{}&{}&{}&{}&{}&{}&{} \
{{color{#4257b2}text{row 1}}}&{}&{}&{}&{}&{}&1&{}&1&{}&{}&{}&{}&{}&{} \
{{color{#4257b2}text{row 2}}}&{}&{}&{}&{}&1&{}&2&{}&1&{}&{}&{}&{}&{} \
{{color{#4257b2}text{row 3}}}&{}&{}&{}&1&{}&3&{}&3&{}&1&{}&{}&{}&{} \
{{color{#4257b2}text{row 4}}}&{}&{}&1&{}&4&{}&6&{}&4&{}&1&{}&{}&{} \
{{color{#4257b2}text{row 5}}}&{}&1&{}&5&{}&{10}&{}&{10}&{}&5&{}&1&{}&{} \
{{color{#4257b2}text{row 6}}}&1&{}&6&{}&{15}&{}&{20}&{}&{15}&{}&6&{}&1&{} \
{}&{}&{}&{}&{}&{}&{}&.&{}&{}&{}&{}&{}&{}&{} \
{}&{}&{}&{}&{}&{}&{}&.&{}&{}&{}&{}&{}&{}&{} \
{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{} \
end{array}
$$
Step 2
2 of 9
$bold{Using;Pascal’s;Triangle;for;Binomial;Expansion:}$ For the binomial expansion
$(a+b)^n$
Perform the following steps
(1) From the exponent $n$, locate the $n^{th}$ row in the Pascal triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.
(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.
In Pascal’s triangle, each row starts with 1. The second number is equal to the row number (called counting number), and third number is a triangular number given by the formula
$S_n=0.5n(n-1)$.
Step 3
3 of 9
[begin{gathered}
{text{a}}{text{.) }},{left( {a + 6} right)^4} hfill \
{text{The numbers at row 4 are 1, 4, 6, 4, 1}} hfill \
hfill \
Rightarrow 1 cdot {a^4} cdot {6^0} hfill \
+ ,4 cdot {a^3} cdot {6^1} hfill \
+ 6 cdot {a^2} cdot {6^2} hfill \
+ 4 cdot {a^1} cdot {6^3} hfill \
+ 1 cdot {a^0} cdot {6^4} hfill \
hfill \
{left( {a + 6} right)^4} = {a^4} + 24{a^3} + 216{a^2} + 864a + 1296 hfill \
end{gathered} ]
{text{a}}{text{.) }},{left( {a + 6} right)^4} hfill \
{text{The numbers at row 4 are 1, 4, 6, 4, 1}} hfill \
hfill \
Rightarrow 1 cdot {a^4} cdot {6^0} hfill \
+ ,4 cdot {a^3} cdot {6^1} hfill \
+ 6 cdot {a^2} cdot {6^2} hfill \
+ 4 cdot {a^1} cdot {6^3} hfill \
+ 1 cdot {a^0} cdot {6^4} hfill \
hfill \
{left( {a + 6} right)^4} = {a^4} + 24{a^3} + 216{a^2} + 864a + 1296 hfill \
end{gathered} ]
Step 4
4 of 9
[begin{gathered}
{text{b}}{text{.) }},{left( {b – 3} right)^5} hfill \
{text{The numbers at row 5 are 1, 5, 10, 10, 5, 1}} hfill \
hfill \
Rightarrow 1 cdot {b^5} cdot {left( { – 3} right)^0} hfill \
+ ,5 cdot {b^4} cdot {left( { – 3} right)^1} hfill \
+ 10 cdot {b^3} cdot {left( { – 3} right)^2} hfill \
+ 10 cdot {b^2} cdot {left( { – 3} right)^3} hfill \
+ 5 cdot {b^1} cdot {left( { – 3} right)^4} hfill \
+ 1 cdot {b^0} cdot {left( { – 3} right)^5} hfill \
hfill \
,{left( {b – 3} right)^5} = {b^5} – 15{b^4} + 90{b^3} + 270{b^2} + 405b – 243 hfill \
end{gathered} ]
{text{b}}{text{.) }},{left( {b – 3} right)^5} hfill \
{text{The numbers at row 5 are 1, 5, 10, 10, 5, 1}} hfill \
hfill \
Rightarrow 1 cdot {b^5} cdot {left( { – 3} right)^0} hfill \
+ ,5 cdot {b^4} cdot {left( { – 3} right)^1} hfill \
+ 10 cdot {b^3} cdot {left( { – 3} right)^2} hfill \
+ 10 cdot {b^2} cdot {left( { – 3} right)^3} hfill \
+ 5 cdot {b^1} cdot {left( { – 3} right)^4} hfill \
+ 1 cdot {b^0} cdot {left( { – 3} right)^5} hfill \
hfill \
,{left( {b – 3} right)^5} = {b^5} – 15{b^4} + 90{b^3} + 270{b^2} + 405b – 243 hfill \
end{gathered} ]
Step 5
5 of 9
[begin{gathered}
{text{c}}{text{.) }},{left( {2c + 5} right)^3} hfill \
{text{The numbers at row 3 are 1, 3, 3, 1}} hfill \
Rightarrow 1 cdot {left( {2c} right)^3} cdot {left( 5 right)^0} hfill \
+ 3 cdot {left( {2c} right)^2} cdot {left( 5 right)^1} hfill \
+ 3 cdot {left( {2c} right)^1} cdot {left( 5 right)^2} hfill \
+ 1 cdot {left( {2c} right)^0} cdot {left( 5 right)^3} hfill \
hfill \
{left( {2c + 5} right)^3} = 8{c^3} + 60{c^2} + 150c + 125 hfill \
end{gathered} ]
{text{c}}{text{.) }},{left( {2c + 5} right)^3} hfill \
{text{The numbers at row 3 are 1, 3, 3, 1}} hfill \
Rightarrow 1 cdot {left( {2c} right)^3} cdot {left( 5 right)^0} hfill \
+ 3 cdot {left( {2c} right)^2} cdot {left( 5 right)^1} hfill \
+ 3 cdot {left( {2c} right)^1} cdot {left( 5 right)^2} hfill \
+ 1 cdot {left( {2c} right)^0} cdot {left( 5 right)^3} hfill \
hfill \
{left( {2c + 5} right)^3} = 8{c^3} + 60{c^2} + 150c + 125 hfill \
end{gathered} ]
Step 6
6 of 9
[begin{gathered}
{text{d}}{text{.) }},{left( {4 – 3d} right)^6} hfill \
{text{The numbers at row 6 are 1, 6, 15, 20, 15, 6, 1}} hfill \
Rightarrow 1 cdot {left( 4 right)^6} cdot {left( { – 3d} right)^0} hfill \
+ ,6 cdot {left( 4 right)^5}{left( { – 3d} right)^1} hfill \
+ 15 cdot {left( 4 right)^4}{left( { – 3d} right)^2} hfill \
+ 20 cdot {left( 4 right)^3}{left( { – 3d} right)^3} hfill \
+ 15 cdot {left( 4 right)^2}{left( { – 3d} right)^4} hfill \
+ 6 cdot {left( 4 right)^1} cdot {left( { – 3d} right)^5} hfill \
+ 1 cdot {left( 4 right)^0} cdot {left( { – 3d} right)^6} hfill \
hfill \
{left( {4 – 3d} right)^6} = 4096 – 18432d + 34560{d^2} – 34560{d^3} + 19440{d^4} – 5832{d^5} + 729{d^6} hfill \
end{gathered} ]
{text{d}}{text{.) }},{left( {4 – 3d} right)^6} hfill \
{text{The numbers at row 6 are 1, 6, 15, 20, 15, 6, 1}} hfill \
Rightarrow 1 cdot {left( 4 right)^6} cdot {left( { – 3d} right)^0} hfill \
+ ,6 cdot {left( 4 right)^5}{left( { – 3d} right)^1} hfill \
+ 15 cdot {left( 4 right)^4}{left( { – 3d} right)^2} hfill \
+ 20 cdot {left( 4 right)^3}{left( { – 3d} right)^3} hfill \
+ 15 cdot {left( 4 right)^2}{left( { – 3d} right)^4} hfill \
+ 6 cdot {left( 4 right)^1} cdot {left( { – 3d} right)^5} hfill \
+ 1 cdot {left( 4 right)^0} cdot {left( { – 3d} right)^6} hfill \
hfill \
{left( {4 – 3d} right)^6} = 4096 – 18432d + 34560{d^2} – 34560{d^3} + 19440{d^4} – 5832{d^5} + 729{d^6} hfill \
end{gathered} ]
Step 7
7 of 9
[begin{gathered}
{text{e}}{text{.) }}{left( {5e – 2f} right)^4} hfill \
{text{The numbers at row 6 are 1, 4, 6, 4, 1}} hfill \
Rightarrow 1 cdot {left( {5e} right)^4}{left( { – 2f} right)^0} hfill \
+ 4 cdot {left( {5e} right)^3} cdot {left( { – 2f} right)^1} hfill \
+ 6 cdot {left( {5e} right)^2} cdot {left( { – 2f} right)^2} hfill \
+ 4 cdot {left( {5e} right)^1} cdot {left( { – 2f} right)^3} hfill \
+ 1 cdot {left( {3e} right)^0} cdot {left( { – 2f} right)^4} hfill \
hfill \
{left( {5e – 2f} right)^4} = 625{e^4} – 1000{e^3}f + 600{e^2}{f^2} – 160e{f^3} + 16{f^4} hfill \
end{gathered} ]
{text{e}}{text{.) }}{left( {5e – 2f} right)^4} hfill \
{text{The numbers at row 6 are 1, 4, 6, 4, 1}} hfill \
Rightarrow 1 cdot {left( {5e} right)^4}{left( { – 2f} right)^0} hfill \
+ 4 cdot {left( {5e} right)^3} cdot {left( { – 2f} right)^1} hfill \
+ 6 cdot {left( {5e} right)^2} cdot {left( { – 2f} right)^2} hfill \
+ 4 cdot {left( {5e} right)^1} cdot {left( { – 2f} right)^3} hfill \
+ 1 cdot {left( {3e} right)^0} cdot {left( { – 2f} right)^4} hfill \
hfill \
{left( {5e – 2f} right)^4} = 625{e^4} – 1000{e^3}f + 600{e^2}{f^2} – 160e{f^3} + 16{f^4} hfill \
end{gathered} ]
Step 8
8 of 9
[begin{gathered}
{text{f}}{text{.) }}{left( {3{f^2} – frac{2}{f}} right)^4} hfill \
{text{The numbers at row 6 are 1, 4, 6, 4, 1}} hfill \
Rightarrow 1 cdot {left( {3{f^2}} right)^4}{left( { – frac{2}{f}} right)^0} hfill \
+ 4 cdot {left( {3{f^2}} right)^3} cdot {left( { – frac{2}{f}} right)^1} hfill \
+ 6 cdot {left( {3{f^2}} right)^2} cdot {left( { – frac{2}{f}} right)^2} hfill \
+ 4 cdot {left( {3{f^2}} right)^1} cdot {left( { – frac{2}{f}} right)^3} hfill \
+ 1 cdot {left( {3{f^2}} right)^0} cdot {left( { – frac{2}{f}} right)^4} hfill \
hfill \
{left( {3{f^2} – frac{2}{f}} right)^4} = 81{f^8} – 216{f^5} + 216{f^2} – 96{f^{ – 1}} + 16{f^{ – 4}} hfill \
end{gathered} ]
{text{f}}{text{.) }}{left( {3{f^2} – frac{2}{f}} right)^4} hfill \
{text{The numbers at row 6 are 1, 4, 6, 4, 1}} hfill \
Rightarrow 1 cdot {left( {3{f^2}} right)^4}{left( { – frac{2}{f}} right)^0} hfill \
+ 4 cdot {left( {3{f^2}} right)^3} cdot {left( { – frac{2}{f}} right)^1} hfill \
+ 6 cdot {left( {3{f^2}} right)^2} cdot {left( { – frac{2}{f}} right)^2} hfill \
+ 4 cdot {left( {3{f^2}} right)^1} cdot {left( { – frac{2}{f}} right)^3} hfill \
+ 1 cdot {left( {3{f^2}} right)^0} cdot {left( { – frac{2}{f}} right)^4} hfill \
hfill \
{left( {3{f^2} – frac{2}{f}} right)^4} = 81{f^8} – 216{f^5} + 216{f^2} – 96{f^{ – 1}} + 16{f^{ – 4}} hfill \
end{gathered} ]
Result
9 of 9
a.) $a^4+24a^3+216a^2+864a+1296$
b.) $b^5-15b^4+90b^3-270b^2+405b-243$
c.) $8c^3+60c^2+150c+125$
d.) $4096-18432d+34560d^2-34560d^3+19440d^4-5832d^5+729d^6$
e.) $625e^4-1000e^3f+600e^2f^2-160ef^3+16f^4$
f.) $81f^8-216f^5+216f^2-96f^{-1}+16f^{-4}$
unlock