$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

Also remember the following terms.

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

begin{table}[]
defarraystretch{2}%
begin{tabular}{|l|l|l|}
hline
mode & $i=dfrac{r}{n}$ & $m=ncdot t$ \ hline
semi-annual & $frac{1}{2}cdot 0.054=0.027$ & $5cdot 2=10$ \ hline
monthly & $frac{1}{12}cdot 0.036=0.003$ & $3cdot 12=36$ \ hline
quarterly & $frac{1}{4}cdot 0.029=0.00725$ & $7cdot 4=28$ \ hline
daily & $frac{1}{52}cdot 0.026=0.0005$ & $frac{10}{12}cdot 52=43frac{1}{3}$ \ hline
end{tabular}
end{table}

begin{table}[]
defarraystretch{2}%
begin{tabular}{|l|l|l|}
hline
mode & $Interest;Rate;per;Compounding;Period$ & $Compounding;Period$ \ hline
semi-annual & $frac{1}{2}cdot 0.054=0.027$ & $5cdot 2=10$ \ hline
monthly & $frac{1}{12}cdot 0.036=0.003$ & $3cdot 12=36$ \ hline
quarterly & $frac{1}{4}cdot 0.029=0.00725$ & $7cdot 4=28$ \ hline
daily & $frac{1}{52}cdot 0.026=0.0005$ & $frac{10}{12}cdot 52=43frac{1}{3}$ \ hline
end{tabular}
end{table}

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

Also remember the following terms.

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

$bf{Solution}$

a.) We shall calculate the accumulated amount for the first 5 years

Since this is compound interest, the common ratio between consecutive compounding period is $left(1+iright)$. In this case, $i=7.2%=0.072$ and compounded annually

1st year: $A(1)=10,000(1.072)=$10;720$

2nd year: $A(2)=10,720(1.072)=$11;491.84$

3rd year: $A(3)=11,491.84(1.072)=$12;319.25$

4th year: $A(4)=12;319.25(1.072)=$13;206.24$

5th year: $A(5)=13;206.24(1.072)=$14;157.09$

The amount after $k$ years,

$A(k)=10;000(1.072)^k$

b.) We shall calculate the accumulated amount for the first 5 half years

Since this is compound interest, the common ratio between consecutive compounding period is $(1+i)$. In this case, $i=1.9%=0.019$ and compounded semi-annually.

1st half-year: $A(1)=10,000(1.019)=$10;190$

2nd half year: $A(2)=10,190(1.019)=$10;383.61$

3rd half year: $A(3)=10,383.61(1.019)=$10;580.90$

4th half year: $A(4)=10;580.90(1.019)=$10;781.94$

5th half year: $A(5)=10;781.94(1.019)=$10;986.80$

The amount after $k$ half-years,

$A(k)=10;000(1.019)^k$

c.) We shall calculate the accumulated amount for the first 5 quarters

Since this is compound interest, the common ratio between consecutive compounding period is $(1+i)$. In this case, $i=1.7%=0.017$ and compounded semi-annually.

1st quarter: $A(1)=10,000(1.017)=$10;170$

2nd quarter: $A(2)=10;170(1.017)=$10;342.89$

3rd quarter: $A(3)=10;342.89(1.017)=$10;518.72$

4th quarter: $A(4)=10;518.72(1.017)=$10;697.54$

5th quarter: $A(5)=10;697.54(1.017)=$10;879.40$

The amount after $k$ quarters,

$A(k)=10;000(1.017)^k$

d.) We shall calculate the accumulated amount for the first 5 quarters

Since this is compound interest, the common ratio between consecutive compounding period is $(1+i)$. In this case, $i=0.9%=0.009$ and compounded semi-annually.

1st month: $A(1)=10,000(1.009)=$10;090$

2nd month: $A(2)=10;090(1.009)=$10;180.81$

3rd month: $A(3)=10;180.81(1.009)=$10;272.44$

4th month: $A(4)=10;272.44(1.009)=$10;364.89$

5th month: $A(5)=10;364.89(1.009)=$10;458.17$

The amount after $k$ months,

$A(k)=10;000(1.009)^k$

a.) $A(k)=10;000(1.072)^k$

b.) $A(k)=10;000(1.019)^k$

c.) $A(k)=10;000(1.017)^k$

d.) $A(k)=10;000(1.009)^k$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

Also remember the following terms.

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

In this case,

$A(t)=P(1+i)^m$

$bold{Solution}$

a.) In this case,

$P=$258$

$r=3.5%=0.035$

$n=1$ (annually)

$A=258(1+0.035)^{10}=$363.93$

b.) In this case,

$P=$5000$

$r=6.4%=0.064$

$n=2$ (semi-annually)

$t=10;years$

$A=5000left(1+dfrac{0.064}{2}right)^{2cdot 10}=$363.93$

c.) In this case,

$P=$12000$

$r=2.8%=0.028$

$n=4$ (quarterly)

$t=6;$ years

$A=1200left(1+dfrac{0.028}{4}right)^{4cdot 6}=$1418.69$

d.) In this case,

$P=$45000$

$r=6%=0.06$

$n=12$ (monthly)

$t=25$ years

$A=45000left(1+dfrac{0.06}{12}right)^{12cdot (25)}=$200923.64$

a.) $$363.93$

b.) $$17;626.17$

c.) $$1418.69$

d.) $$200;923.64$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

Also remember the following terms.

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A(t)-P=P[(1+i)^m-1]
$$

$bold{Solution}$

a.) We are given with

$P=4000$

$i=0.03$

$m=4$

Using the formula above,

$A=P(1+i)^m=4000(1+0.03)^4=$4502$

$$
I=A-P=4502.04-4000=$502.04
$$

b.) We are given with

$P=7500$

$i=0.005$

$m=72$

Using the formula above,

$A=P(1+i)^m=7500(1+0.005)^{72}=$10;740.33$

$$
I=A-P=10740.33-7500=$3240.33
$$

c.) We are given with

$P=15000$

$i=0.006$

$m=20$

Using the formula above,

$A=P(1+i)^m=15000(1+0.006)^{20}=$16;906.39$

$$
I=A-P=16906.39-15000=$1906.39
$$

d.) We are given with

$P=28200$

$i=0.0275$

$m=20$

Using the formula above,

$A=P(1+i)^m=15000(1+0.006)^{20}=$48;516.08$

$$
I=A-P=48;516.08-28;200=$20;316.08
$$

e.) We are given with

$P=850$

$i=0.0001$

$m=365$

Using the formula above,

$A=P(1+i)^m=850(1+0.0001)^{365}=$881.60$

$$
I=A-P=881.60-850=$31.60
$$

f.) We are given with

$P=2225$

$i=0.001$

$m=47$

Using the formula above,

$A=P(1+i)^m=2225(1+0.001)^{47}=$2332.02$

$$
I=A-P=2332.02-2225=$107.02
$$

a.) $A=$4502.04$ ; $I=$502.04$

b.) $A=$10;740.33$ ; $I=$3240.33$

c.) $A=$16;906.39$ ; $I=$1906.39$

d.) $A=$48;516.08$ ; $I=$20;316.08$

e.) $A=$881.60$ ; $I=$31.60$

d.) $A=$2332.02$ ; $I=$107.02$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

a.) Remember that ratio of $A$ between consecutive compounding is $left(1+dfrac{r}{n}right)$. For annual compounding, $n=1$, so this ratio becomes $(1+r)$

$$
dfrac{4404.404240}{4240}=1+rimplies r=dfrac{4404.404240}{4240}-1=0.06=boxed{bold{6%}}
$$

b.) Here we shall find $P$ given that after first year, $A=4240$

$$
A=Pleft(1+dfrac{0.06}{1}right)^{1(1)}implies P=dfrac{4240}{1+0.06}=boxed{bold{$4000}}
$$

a.) $6%$

b.) $$4,000$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

Here we are given with $A=25;000$, $r=0.07$, $P=10;000$, and monthly compounding (n=12).

We shall find how long will it take to reach $$25;000$.

Using the formula

$A=Pleft(1+dfrac{r}{n}right)^m$

$A=Pleft(1+dfrac{0.07}{12}right)^{m}$

$25,000=10,000left(1+dfrac{0.07}{12}right)^mimplies 2.5=left(1+dfrac{0.07}{12}right)^m$

We shall try several values of $m$

if $m=156implies left(1+dfrac{0.07}{12}right)^{156}=2.4778$

if $m=157implies left(1+dfrac{0.07}{12}right)^{157}=2.4922$

if $m=158implies left(1+dfrac{0.07}{12}right)^{158}=2.5068$

Thus, the account value will be at least $25,000$ at the 158$^{th}$ compounding.

Since compounding is done every month, this is also $boxed{bold{158; months}}$.

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

We shall find which one is a cheaper option for $P=15,000$

Option 1:

$r=0.1$

$n=4$ (quarterly)

$t=10;text{years}$

Option 2:

for first 5 years:
$r=0.12$

$n=4$

$t=5;text{years}$

for next 5 years:
$r=0.06$

$n=4$

$t=5;text{years}$

Option 1 :
$A=15,000cdot left( 1+dfrac{0.1}{4}right)^{4(10)}=underline{$40,275.96}$

Option 2:
$A_1=15,000cdot left( 1+dfrac{0.12}{4}right)^{4(5)}=$27,091.67$

$A_2=27,091.67cdot left(1+dfrac{0.06}{4}right)^{4(5)}=$36;488.55$

Total = $$27,091.67+$36,488.55=underline{$36,488.55} ;;checkmark$

Therefore, Option 2 is cheaper by $40275.96-36488.55=$3787.41$

Choose option 2 and save $$3787.41$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

By inspection, you can see that in the formula

$A(n)=5000times (1.0075)^{12n}$

$P=$5000$

$i=0.0075$

$n=12$ (monthly compounding)

$r=icdot n=0.0075cdot 12=0.09=9%$ per annum

$P=$5000$ , monthly compounding, $r=9%$ annual interest rate

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

We shall evaluate which payment plan is cheaper for $P=949.99$

Plan A: simple interest, $r=0.10$ for 2 years

$A=P(1+rt)$

$A_A=949.99(1+0.1cdot 2)=boxed{$1399.99}$

Plan B: compound interest, $i=0.0125$, $m=8$

$A=P(1+i)^m$

$A_B=949.99(1+0.0125)^8=boxed{$1049.25};;checkmark$

Plan B is cheaper by $1399.99-1049.25=$290.64$

Choose plan B to save $$290.64$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

In this case, we are given with $P=1000$, $i=5%=0.05$, and $m=7$.

We can determine the future value after 7 compounding using this formula:

$A=P(1+i)^m$

$A=1000(1+0.05)^7=boxed{bold{$1407.10}}$

$$
$1407.10
$$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

In this case, we are given with

for first 3 years: $P=9000$ ; $r=10%=0.1$ ; $n=4$
for next 2 years: $r=9%=0.09$ ; $n=2$

Using the formula, the amount after 3 years is

$A_1=Pleft(1+dfrac{r}{n}right)^{nt}=9000left(1+dfrac{0.1}{4}right)^{4cdot 3}=9000left(1+dfrac{0.1}{4}right)^{4(3)}=$12;104.00$

The amount after 3 years will be the principal for the next two years

$A_2=12104.00left(1+dfrac{0.09}{2}right)^{2(2)}=$14;434.24$

The amount after 5 years is $boxed{bold{$14;434.24}}$

$$
$14;434.24
$$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

We need to how many years will a simple interest with annual rate of $10%$ would have equal value with a compound interest with $5%$ annually.

$Pleft(1+rtright)=Pcdot (1+r)^{t}$

$1+0.05t=(1+0.05)^{t}$

This equation is difficult to solve analytically, so we will solve it graphically.

In the graph, the red curve is the simple interest while the blue curve is the compound interest. This indicates that before $tapprox 26.583$ years, simple interest is more profitable, but after 26.583 years, the compound interest gives more profits.

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

We have to cases:

(a) $$5000$ at $5%$ compounded annually

(b) $$3000$ at $7%$ compounded annually

$5000cdot (1.05)^t=3000(1.07)^t$

$left(dfrac{1.07}{1.05}right)^t=dfrac{5000}{3000}=1.66667$

We shall find value $t$ by taking several values of $t$

$t=26implies left(dfrac{1.07}{1.05}right)^{26}approx 1.6333$

$t=27implies left(dfrac{1.07}{1.05}right)^{27}approx 1.6644$

$t=28implies left(dfrac{1.07}{1.05}right)^{28}approx 1.6961$

Thus, the two cases would have equivalent value at around $boxed{bold{ 27 years}}$.

$27$ years

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bold{Solution}$

We will calculate the future worth of $P=$1$ after 1 year to compare which investment plan is more profitable.

Case 1: $6.5%$ quarterly

$A=1cdotleft(1+dfrac{0.065}{4}right)^4approx 1.066602$

Case 2: $6.55%$ semi-annually

$A=1cdot left(1+dfrac{0.0655}{2}right)^2approx 1.066572$

Case 3: $6.45%$ monthly

$A=1cdot left(1+dfrac{0.0645}{2}right)^{12}approx 1.066441$

Case 4: $6.6%$ annually

$A=1cdot left(1+dfrac{0.066}{1}right)^{1}=1.066$

Therefore, $6.5%$ compounded quarterly is more profitable.

$6.5%$ compounded quarterly is more profitable.

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bf{Solution}$

We shall find the amount on July 1, 2001:

$P=2000$ ; $i=0.5%=0.005$, $m=60$

$A_1=2000(1+0.005)^{60}=$2697.70$

The new principal for the next 26 compounding is $$2697.70$

$P=2697.70$ , $i=0.02$, $m=26$

$A_2=2697.70(1+0.02)^{26}=$4514.48$

Therefore, the account value on the first day of 2008 is $boxed{bold{$4,514.38}}$

$$
$4,514.38
$$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bf{Solution}$

We’re given $P=4000$, $r=4%=0.04$, compounded quarterly (n=4).

The amount after 1 year

$A=Pleft(1+dfrac{r}{n}right)^{nt}=4000left(1+dfrac{0.04}{4}right)^{4(1)}=$4162.42$

For the second year $r=4.2%=0.042$

$A=4162.42left(1+dfrac{0.042}{4}right)^{4(1)}=$4340.01$

For the third year: $r=4.4%=0.044$

$A=4340.01left(1+dfrac{0.044}{4}right)^{4(1)}=$4534.14$

The amount after third year is $boxed{bold{$4534.14}}$

$$
$4534.14
$$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bf{Solution}$

For the first five years, $P=500$, $r=4.8%=0.048$, $n=12$ (monthly compoundly)

$A_1=500left(1+dfrac{0.048}{12}right)^{12(5)}=$635.32$

On her 5th birthday, she received another $$500$, so the amount on her $10^{th}$ birthday is

$A_2=(635.32+500)left(1+dfrac{0.048}{12}right)^{12(5)}=$1442.58$

On her 10th birthday, she received another $$500$, so the amount on her $15^{th}$ birthday is

$A_3=(1442.58+500)left(1+dfrac{0.048}{12}right)^{12(5)}=$2468.32$

On her 15th birthday, she received another $$500$, so the amount on her $18^{th}$ birthday is

$A_4=(2468.32+500)left(1+dfrac{0.048}{12}right)^{12(3)}=$3427.08$

The account value on her 18th birthday is $boxed{bold{$3427.08}}$.

$$
$3427.08
$$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bf{Solution}$

With semi-annual compounding, there will be $20$ compounding period within 10 years. Every period, Anita adds $$500$ from an initial investment $P=$500$, $n=2$, $r=6.8%=0.068$, $i=frac{r}{n}=frac{0.068}{2}=0.034$ and $t=10;$ years.
The principal $$500$ will be compounded $10times 2=20$ times, thus

$A_0=500(1+0.034)^{20}$

The first added $$500$ will be compounded 19 times,

$A_1=500(1+0.034)^{19}$

The second added $$500$ will be compounded 18 times

$A_2=500(1+0.034)^{18}$

Thus, the amount due to the $k^{th}$ investment is

$A_k=500(1+0.034)^{20-k}$

The total investment at 20th compounding is then a geometric series with

$t_1=500(1.034)^{19}=943.76$

$r=dfrac{500(1.034)^{19}}{500(1.034)^{20}}=dfrac{1}{1.034}$

$S_{20}=t_1cdot dfrac{r^n-1}{r-1}=943.76cdot dfrac{(1/1.034)^{20}-1}{(1/1.034)-1}=13995.48$

Therefore, after 10 years, the account value would be $boxed{bold{$13,995.48}}$.

We can confirm our answer using the standard formula for the future value of compounded annuity.

$$
A=Pcdot dfrac{(1+i)^m-1}{i}=500cdot dfrac{(1+0.034)^{20}-1}{0.034}=$13995.48
$$

$$
$13,995.48
$$

$bold{Compound;Interest;Formula}$

$A(t)=Pleft(1+dfrac{r}{n}right)^{nt}$

where

$A(t)$ = future value of investment or debt after $t$ years

$P$ = principal or initial deposit or debt

$r$ = annual interest rate

$n$ = number of compounding per year ($n=1$ for annually, $n=2$ for semi-annually, $n=4$ for quarterly, $n=12$ for monthly)

$t$ = number of years that the money is invested or borrowed

$bold{Useful;Definitions}$

Interest Rate per Compounding Period: $i=dfrac{r}{n}$

Number of Compounding Periods: $m=ncdot t$

Interest: $I=A-P$

Using this variables, the formula becomes,

$A=P(1+i)^m$

$$
I=A-P=P[(1+i)^m-1]
$$

$bf{Solution}$

a.) We shall find the effective annual interest rate for a semi-annual compounding (n=2) with $r=6.3%$.

$A=Pleft(1+dfrac{r}{n}right)^{nt}$

$Pleft(1+iright)^{1}=Pleft(1+dfrac{0.063}{2}right)^{2}$

Divide both sides by $P$ and solve for $i$

$$
i=left(1+dfrac{0.063}{2}right)^{2}-1=0.064=6.4%
$$

b.) We shall find the effective annual interest rate for a monthly compounding with $r=4.2%$.

$Pleft(1+iright)^{1}=Pleft(1+dfrac{0.042}{12}right)^{12}$

$$
i=left(1+dfrac{0.042}{12}right)^{12}-1=0.0428=4.28%
$$

b.) We shall find the effective annual interest rate for a quarterly compounding with $r=3.2%$.

$Pleft(1+iright)^{1}=Pleft(1+dfrac{0.032}{4}right)^{4}$

$$
i=left(1+dfrac{0.032}{4}right)^{4}-1=0.0324=3.24%
$$

a.) $6.40%$

b.) $4.28%$

c.) $3.24%$