Chapter 12, Problem 3P

Monthly Savings Program Alice opens a savings account that pays 3% interest per year, compounded monthly. She begins by depositing $100 at the start of the first month and adds $100 at the end of each month, when the interest is credited.

1. (a) Find a recursive formula for the amount A_n in her account at the end of the nth month. (Include the interest credited for that month and her monthly deposit.)
2. (b) Find the first five terms of the sequence A_n.
3. (c) Use the pattern you observed in (b) to find a formula for A_n. [Hint: To find the pattern most easily, it’s best not to simplify the terms too much]
4. (d) How much has she saved after 5 years?

To determine

The recursive formula for the amount $A_n$ in the account at the end of the $n\text{th}$ month.

Answer to Problem 3P

The recursive formula for the amount $A_n$ in the account at the end of the $n\text{th}$ month is $A_n=A_{n-1}\left(1.0025\right)+100$.

Explanation of Solution

Given:

Alice deposit $100 at the start of first month and then add $100 at the end of each month. The interest rate is 3% per year and interest compounded monthly.

Formula used:

The compound interest formula in given below,

$A=P{\left(1+\frac{r}{n}\right)}^{nt}$ (1)

Where, A is total amount after t years, r is the rate of interest per year and n is no of times interest is compounded per year

Calculation:

According to the given information the value of r is 3% or 0.03 and the value of n is 12. Alice add $100 at the end of the each month, so the amount after one month is calculated as follows,

Substitute 0.03 for r, 12 for n, 100 for P and $\frac{1}{12}$ for t is equation (1) to find the total amount in first month after compounding the interest.

$\begin{array}{c}A=100{\left(1+\frac{0.03}{12}\right)}^{\left(12\right)\left(\frac{1}{12}\right)}\\ =100{\left(1+0.0025\right)}^1\\ =100\left(1.0025\right)\\ =0.25\end{array}$

Add 100 and the total amount in first month after compounding the interest, to find the total amount after first month.

$\begin{array}{c}{A}_1=0.25+100\\ =100.25\end{array}$

The new principle amount is $100.25$.

Substitute 0.03 for r, 12 for n, $A_1$ for P and $\frac{1}{12}$ for t is equation (1) to find the total amount in second month after compounding the interest.

$\begin{array}{c}A=A_1{\left(1+\frac{0.03}{12}\right)}^{\left(12\right)\left(\frac{1}{12}\right)}\\ =A_1{\left(1+0.0025\right)}^1\\ =A_1\left(1.0025\right)\end{array}$

Add 100 and the total amount in second month after compounding the interest, to find the total amount after second month.

$A_2=A_1\left(1.0025\right)+100$

The new principle amount is $A_1\left(1.0025\right)+100$.

Similarly, the principle amount after n month is $A_{n-1}$ and Alice add $100 in end of the month, so the total amount after n month is,

$A_n=A_{n-1}\left(1.0025\right)+100$.

Therefore, the recursive formula for the amount $A_n$ in the account at the end of the $n\text{th}$ month is $A_n=A_{n-1}\left(1.0025\right)+100$.

To determine

The first five terms of the sequence $A_n$.

Answer to Problem 3P

The first five terms of the sequence $A_n$ are $A_0=100$, $A_1=200.25$, $A_2=300.75$, $A_3=401.50$ and $A_4=502.50$.

Explanation of Solution

Given:

From part (a), the value of $A_n$ is given below,

$A_n=A_{n-1}\left(1.0025\right)+100$ (2)

Calculation:

The initial amount is $100. The value of first term $A_0$ is 100.

Substitute 1 for n in equation (2), to find the value of $A_1$.

$\begin{array}{c}{A}_1=A_{1-1}\left(1.0025\right)+100\\ =A_0\left(1.0025\right)+100\\ =100\left(1.0025\right)+100\left(\because A_0=100\right)\end{array}$

Simplify the above expression to find the value of $A_1$.

$\begin{array}{c}{A}_1=100.25+100\\ =200.25\end{array}$

Substitute 2 for n in equation (2), to find the value of $A_2$.

$\begin{array}{c}{A}_2=A_{2-1}\left(1.0025\right)+100\\ =A_1\left(1.0025\right)+100\\ =\left(200.25\right)\left(1.0025\right)+100\left(\because A_1=200.25\right)\end{array}$

Simplify the above expression to find the value of $A_2$.

$\begin{array}{c}{A}_2=200.75+100\\ =300.75\end{array}$

Substitute 3 for n in equation (2), to find the value of $A_3$.

$\begin{array}{c}{A}_3=A_{3-1}\left(1.0025\right)+100\\ =A_2\left(1.0025\right)+100\\ =\left(300.75\right)\left(1.0025\right)+100\left(\because A_2=300.75\right)\end{array}$

Simplify the above expression to find the value of $A_3$.

$\begin{array}{c}{A}_3=301.50+100\\ =401.50\end{array}$

Substitute 4 for n in equation (2), to find the value of $A_4$.

$\begin{array}{c}{A}_4=A_{4-1}\left(1.0025\right)+100\\ =A_3\left(1.0025\right)+100\\ =\left(401.50\right)\left(1.0025\right)+100\left(\because A_3=401.50\right)\end{array}$

Simplify the above expression to find the value of $A_4$.

$\begin{array}{c}{A}_4=402.50+100\\ =502.50\end{array}$

Therefore, the first five terms of the sequence $A_n$ are $A_0=100$, $A_1=200.25$, $A_2=300.75$, $A_3=401.50$ and $A_4=502.50$.

To determine

The formula for $A_n$ using the pattern of in part (b).

Answer to Problem 3P

The formula for $A_n$ using the pattern of in part (b) is $100\left(\frac{{\left(1.0025\right)}^{n+1}-1}{0.0025}\right)$.

Explanation of Solution

Given:

The first five terms of the sequence $A_n$ are $A_0=100$, $A_1=200.25$, $A_2=300.75$, $A_3=401.50$ and $A_4=502.50$.

Calculation:

The first five terms represents as shown below,

$\begin{array}{c}{A}_0=100\\ =100\left(\frac{{\left(1.0025\right)}^{0+1}-1}{0.0025}\right)\end{array}$

The second term is written as,

$\begin{array}{c}{A}_1=200.25\\ =100\left(\frac{{\left(1.0025\right)}^{1+1}-1}{0.0025}\right)\end{array}$

The third term is written as,

$\begin{array}{c}{A}_2=300.75\\ =100\left(\frac{{\left(1.0025\right)}^{2+1}-1}{0.0025}\right)\end{array}$

The fourth term is written as,

$\begin{array}{c}{A}_3=401.50\\ =100\left(\frac{{\left(1.0025\right)}^{3+1}-1}{0.0025}\right)\end{array}$

The fifth term is written as,

$\begin{array}{c}{A}_5=502.50\\ =100\left(\frac{{\left(1.0025\right)}^{4+1}-1}{0.0025}\right)\end{array}$

Similarly the general formula for $A_n$ is written as shown below,

$\begin{array}{c}{A}_n=502.50\\ =100\left(\frac{{\left(1.0025\right)}^{n+1}-1}{0.0025}\right)\end{array}$

Therefore, the formula for $A_n$ using the pattern of in part (b) is $100\left(\frac{{\left(1.0025\right)}^{n+1}-1}{0.0025}\right)$.

To determine

The total amount saved by Alice in 5 years.

Answer to Problem 3P

The total amount saved by Alice in 5 years is $6580.83.

Explanation of Solution

Given:

From part a the formula for $A_n$ is given below,

$A_n=100\left(\frac{{\left(1.0025\right)}^{n+1}-1}{0.0025}\right)$ (3)

Where, $A_n$ is total saved amount after n months.

Calculation:

The total number of months in 5 years is 60.

Substitute 60 for n in equation (3), to find the value of total saved amount after 5 years.

$\begin{array}{c}{A}_{60}=100\left(\frac{{\left(1.0025\right)}^{60+1}-1}{0.0025}\right)\\ =100\left(\frac{1.16-1}{0.0025}\right)\\ =100\left(\frac{0.16}{0.0025}\right)\end{array}$

Simplify the above expression,

$\begin{array}{c}{A}_{60}=100\left(65.8083\right)\\ =6580.83\end{array}$

Therefore, the total amount saved by Alice in 5 years is $6580.83.