We can observe that the recursive formula appears valid across all terms and considers only the two terms preceding it. Notice also that the first 6 terms are getting repeated periodically implying that

$t_1=t_7=1$

$t_2=t_8=5$

$t_3=t_9=4$

$t_4=t_{10}=-1$

$t_5=t_{11}=-5$

$t_n=t_{n+6}$

Observe that the numerators N form an arithmetic sequence with $t_1=1$ and $d=1$. This can be written as

$N=1+(n-1)(1)=n$

The denominators D form an arithmetic sequence with $t_1=2$ and $d=1$.

This can be written as

$D=2+(n-1)(1)=n+1$

Therefore, the original sequence has a general term of

$$
t_n=dfrac{N}{D}=dfrac{n}{n+1}
$$

$$
t_n=dfrac{n}{n+1}
$$

a.) A triangle has 3 sides so 3 toothpicks is needed for the first triangle. To form a new triangle adjacent to it, we only need 2 toothpicks since the 3rd side is already on the triangle preceding it. Thus, this is an arithmetic sequence with $t_1=3$ and $d=2$, and the general term is

$t_n=t_1+(n-1)d$

$t_n=3+(n-1)(2)$

$$
t_n=2n+1
$$

b.) A square has four sides, so the first one should require 4 toothpicks ,that is, $t_1=4$. The new square adjacent to it then requires 3 more since one side is already present from the existing ones, thus, $d=3$. The general term is

$$
t_n=4+(n-1)(3)=3n+1
$$

c.) To form an $4times4$ grid, we need 5 rows with 4 horizontal toothpicks each, to make $ntimes n$ grid, we need $(n+1)(n)=n^2+n$ horizontal tooth picks. The number of vertical toothpicks is equal to that of horizontal ones, so the total number is

$$
t_n=2(n^2+n)=2n^2+2n
$$

d.) For $n=4$,

for a row of 4 triangles: $3(4)+1=12$

for $4times 4$ square grid: $2(4)^2+2(4)=40$

a.) $t_n=2n+1$

b.) $t_n=3n+1$

c.) $t_n=2n^2+2n$

d.) 12 ; 40

From the given sequence, we can observe the following:

(1) When $n$ is even, $t_n$ is positive, and $t_n$ is half of $n$.

(2) When $n$ is odd, $t_n$ is non-positive, and $t_n$ equals $-t_{n-1}$.

$n=12;345$ is odd

so it will follow $t_n=-dfrac{1}{2}(n-1)=-dfrac{1}{2}(12;345-1)=-6172$

From the given sequence, we can observe the following:

(1) The first term appears to be arithmetic with $a_1=x$ and $d=x$

Thus, $a_n=x+(n-1)(x)=nx$

(2) The second term appears to be geometric with $g_1=dfrac{1}{y}$ and $r=dfrac{1}{y}$

Thus, $g_n=dfrac{1}{y}cdot left(dfrac{1}{y}right)^{n-1}=dfrac{1}{y^n}$

Therefore, the original sequence can be written as

$t_n=a_n+g_n$

$$
t_n=nx+dfrac{1}{y^n}
$$

$$
t_n=nx+dfrac{1}{y^n}
$$

From the given sequence, we can observe the following:

(1) The numerator appears to be geometric with $t_1=3$ and $r=dfrac{21}{3}=7$

Thus, $t_n=3cdot 7^{n-1}$

(2) The denominator appears to increasing repeating digits of 5. To obtain the general term, we must examine the origin of the repeating digits. Observe that

$dfrac{5}{9}cdot 9=5$

$dfrac{5}{9}cdot 99=55$

$dfrac{5}{9}cdot 999=555$

Meanwhile,

$9=10-1$

$99=100-1$

$999=1000-1$

Thus, we can see a pattern that $t_n=dfrac{5}{9}(10^n-1)$.

Therefore, the original sequence has a general term of

$$
t_n=dfrac{3cdot 7^{n-1}}{dfrac{5}{9}(10^n-1)}
$$

$$
t_n=dfrac{3cdot 7^{n-1}}{dfrac{5}{9}(10^n-1)}
$$

a.) $4,9,19,39,79$

Notice that

$9=2(4)+1$

$19=2(9)+1$

$39=2(19)+1$

$t_n=2cdot t_{n-1}+1$

Using this recursive formula, we can solve the next three terms as

$t_6=2(79)+1=159$

$t_7=2(159)+1=319$

$$
t_8=2(319)+1=639
$$

b.) The first differences is an arithmetic sequence with $d=-1$

Knowing that $t_5=90$ and $t_6-t_5=-5$

$t_6=90-5=85$

$t_7=85-6=79$

$$
t_8=79-7=72
$$

c.) Find the difference between consecutive terms. The first differences $0,1,1,2,3,5,8…$ appears to be $d_n=d_{n-1}+d_{n-2}$ for $n>2$

Knowing that $t_8=21$ and $t_7=13$

$t_9=21+13=34$

$t_{10}=34+21=55$

$$
t_{11}=55+34=89
$$

d.) Find the difference between consecutive terms. Notice that it forms a sequence

$2,;5,;2,;12,;2,;26$

Observe that when $n$ is even, the difference between the term preceding it is 2, thus,

for even $n$, $t_n=t_{n-1}+2$

If $n$ is odd, $t_n$ is two times the term preceding it, this means,

for odd $n$, $t_n=2cdot t_{n-1}$

Knowing that $t_7=52$, we can obtain the next three terms as

$t_8=t_7+2=52+2=54$

$t_9=2cdot(54)=108$

$$
t_{10}=108+2=110
$$

e.) Observe that all the terms are perfect cubes of $n$, but for even $n$, $t_n$ is negative, while for odd $n$, $t_n$ positive. Thus, the next three terms are:

$t_6=-6^3=-216$

$t_7=7^3=343$

$$
t_8=-8^3=-512
$$

f.) Observe that

$t_1=6$

$t_2=2(6)+1=13$

$t_3=2(13)+1=27$

$t_4=2(27)+1=55$

Thus, the recursive formula is $t_n=2cdot2t_{n-1}+1$

We can then calculate the next three terms as

$t_5=2(55)+1=111$

$t_6=2(111)+1=223$

$$
t_7=2(223)+1=447
$$

We shall a find a pattern starting from smaller sequences following the algorithm described.

For 2 ,1

Compare 2 and 1 $implies$ 1,2

Compare 1 and 2 $implies$ Leave as is

A sequence of 2 numbers requires 2 comparisons.

For 3, 2, 1

Compare 3 and 2 $implies$ 2, 3, 1

Compare 3 and 1 $implies$ 2, 1, 3

Compare 2 and 1 $implies$ 1, 2, 3

Compare 2 and 3 $implies$ Leave as is

Compare 1 and 2 $implies$ Leave as is

Compare 2 and 3 $implies$ Leave as is

A sequence of 3 numbers requires 6 comparisons

For 4, 3, 2, 1

Compare 4 and 3 $implies$ 3, 4, 2 ,1

Compare 4 and 2 $implies$ 3, 2, 4, 1

Compare 4 and 1 $implies$ 3, 2, 1, 4

Compare 3 and 2 $implies$ 2, 3, 1, 4

Compare 3 and 1 $implies$ 2, 1, 3, 4

Compare 3 and 4 $implies$ Leave as is

Compare 2 and 1 $implies$ 1, 2, 3, 4

Compare 2 and 3 $implies$ Leave as is

Compare 3 and 4 $implies$ Leave as is

Compare 1 and 2 $implies$ Leave as is

Compare 2 and 3 $implies$ Leave as is

Compare 3 and 4 $implies$ Leave as is

A sequence of 4 numbers requires 12 comparisons

Let’s start by the sequence 5, 4, 3, 2, 1

The algorithm would be

Compare 5 and 4 $implies$ 4, 5, 3, 2, 1

Compare 5 and 3 $implies$ 4, 3, 5, 2, 1

Compare 5 and 2 $implies$ 4, 3, 2, 5, 1

Compare 5 and 1 $implies$ 4, 3, 2, 1, 5

Compare 4 and 3 $implies$ 3, 4, 2, 1, 5

Compare 4 and 2 $implies$ 3, 2, 4, 1, 5

Compare 4 and 1 $implies$ 3, 2, 1, 4, 5

Compare 4 and 5 $implies$ Leave as is

Compare 3 and 2 $implies$ 2, 3, 1, 4, 5

Compare 3 and 1 $implies$ 2, 1, 3, 4, 5

Compare 3 and 4 $implies$ Leave as is

Compare 4 and 5 $implies$ Leave as is

Compare 2 and 1 $implies$ 1, 2, 3, 4 ,5

Then the algorithm would make 7 more comparison with no changes and stop.

So to sort 5 numbers requires 20 comparisons.

The sequence describing the number of comparisons to the number of terms in the sequence is

2, 6, 12, 20

Neither common difference nor common ratio exists so we must find another pattern.

Observe that

$n = 2 implies 2times 1= 2$

$n = 3 implies 3times 2= 6$

$n = 4 implies 4times 3= 12$

$n = 5 implies 5times 4= 20$

We see that the pattern is
$t_n=n(n-1)$

Therefore, for $n=100$, the number of comparisons for this algorithm is

$$
t_{100}=100(100-1)=9900
$$

$9,900$ comparisons

By inspection, observe some pattern. Notice that

$t_1=2$

$t_2=5(2)+1=11$

$t_3=5(11)-1=54$

$t_4=5(54)+1=271$

$t_5=5(271)-1=1354$

$t_6=5(1354)+1=6771$

Therefore, the next 3 terms are

$t_7=5(6771)-1=33854$

$t_8=5(33;854)+1=169;271$

$t_9=5(169;271)-1=846;354$

$$
t_{10}=5(846354)+1=4;231;771
$$

From the given recursive formula, we shall observe some pattern by solving for the first few terms.

$t_1=1$

$t_2=dfrac{5}{2}(1+1)=5$

$t_3=dfrac{5}{2}(5+1)=15$

$t_4=dfrac{5}{2}(15+1)=40$

$t_5=dfrac{1}{2}(40)=20$

$t_6=dfrac{1}{2}(20)=10$

$t_7=dfrac{1}{2}(10)=5$

$$
t_8=dfrac{5}{2}(5+1)=15
$$

Observe that the cycle repeats every 5 terms, for example $t_2=t_5$, $t_3=t_8$. Thus, any $n$ divisible by 5 will take a value of $20$, such as $t_{55}, t_{95}, t_{995}$.

Therefore $t_{1000}=20$

$$
t_{1000}=20
$$

Examples of sequences that are neither arithmetic nor geometric are those that are periodic (undergoes cycles) by providing a rule for even or odd $n$.

For instance, $3,;7,;3,;7,;3,;7…$, the general term would be

$t_n= 3$ if $n$ is odd

$t_n=7$ if $n$ is even.

Thus, sequences like

$1,;2,;1,;2,;1,;2,;1$

or

$5,;10,;5,;10,;5,;10$

would fit this description.