Chapter 12.1, Problem 78E

(a)

To determine

To find: First six terms of amount of interest generated.

Answer to Problem 78E

The first 6 terms of amount of interest generated are $\$0,\$0.5,\$1.5,\$3.01,\$5.03,\$7.55$.

Explanation of Solution

Given:

Principal amount deposited in account is $\$100$,

Rate of interest compounded monthly is $6\%$,

The amount of interest in account after n months is,

$I_n=100\left(\frac{{1.005}^n-1}{0.005}-n\right)$ (1)

Calculation:

Substitute 1 for n in equation (1) to calculate amount after first month,

$\begin{array}{c}{I}_1=100\left(\frac{{1.005}^1-1}{0.005}-1\right)\\ =0\end{array}$

Substitute 2 for n in equation (1) to calculate amount after second month,

$\begin{array}{c}{I}_2=100\left(\frac{{1.005}^2-1}{0.005}-2\right)\\ =0.5\end{array}$

Substitute 3 for n in equation (1) to calculate amount after third month,

$\begin{array}{c}{I}_3=100\left(\frac{{1.005}^3-1}{0.005}-3\right)\\ =1.5\end{array}$

Substitute 4 for n in equation (1) to calculate amount after fourth month,

$\begin{array}{c}{I}_4=100\left(\frac{{1.005}^4-1}{0.005}-4\right)\\ =3.01\end{array}$

Substitute 5 to for n in equation (1) to calculate amount after fifth month,

$\begin{array}{c}{I}_5=100\left(\frac{{1.005}^5-1}{0.005}-5\right)\\ =5.03\end{array}$

Substitute 6 for n in equation (1) to calculate amount after sixth month,

$\begin{array}{c}{I}_6=100\left(\frac{{1.005}^6-1}{0.005}-6\right)\\ =7.55\end{array}$

Thus the first 6 terms of amount of interest generated are $\$0,\$0.5,\$1.5,\$3.01,\$5.03,\$7.55$.

(b)

To determine

To find: Amount in account after 3 years.

Answer to Problem 78E

The amount in account after 3 years are $\$2149.16$.

Explanation of Solution

Given:

Principal amount deposited in account is $\$100$,

Rate of interest compounded monthly is $6\%$,

The amount of interest in account after n months is,

$I=100\left(\frac{{1.005}^n-1}{0.005}-n\right)$ (1)

Calculation:

Substitute $60$ for n in equation (1) to calculate amount after 5 years,

$\begin{array}{l}I=100\left(\frac{{1.005}^{60}-1}{0.005}-60\right)\\ =977.00\end{array}$

Thus the amount in account after 3 years are $\$977.00$.