if $11,500 per quarter is invested in an account that earns a nominal annual interest rate of 13% compounded…

if $11,500 per quarter is invested in an account that earns a nominal annual interest rate of 13% compounded monthly, how much money will be in the account at the end of 4 years? the account will have $ (round to the nearest dollar.)

if $11,500 per quarter is invested in an account that earns a nominal annual interest rate of 13% compounded monthly, how much money will be in the account at the end of 4 years? the account will have $ (round to the nearest dollar.)

Answer

Explanation:

Step1: Determine the number of compounding periods and the interest rate per period

The nominal annual interest rate $r = 13%=0.13$, and it is compounded monthly. So the monthly - interest rate $i=\frac{r}{12}=\frac{0.13}{12}$. The investment is made quarterly for $n = 4$ years. Since there are 4 quarters in a year, the number of quarterly - payments $m=4\times4 = 16$.

Step2: First, find the effective quarterly interest rate

Let the effective quarterly interest rate be $j$. We use the formula $(1 + j)=(1+\frac{0.13}{12})^3$ (because there are 3 months in a quarter). [1 + j=(1+\frac{0.13}{12})^3=1+\frac{0.13}{4}+\frac{3\times(\frac{0.13}{12})^2}{1}+\frac{(\frac{0.13}{12})^3}{1}\approx1 + 0.0325+0.00027+0.000003\approx1.03277]

Step3: Use the future - value of an ordinary annuity formula

The formula for the future value of an ordinary annuity is $F = A\times\frac{(1 + j)^m-1}{j}$, where $A = 11500$, $j\approx0.03277$, and $m = 16$. [F=11500\times\frac{(1.03277)^{16}-1}{0.03277}] First, calculate $(1.03277)^{16}$. Using the formula $a^n=e^{n\ln(a)}$, we have $\ln(1.03277)\approx0.0322$ and $n = 16$, so $(1.03277)^{16}=e^{16\times0.0322}=e^{0.5152}\approx1.674$. [F = 11500\times\frac{1.674 - 1}{0.03277}=11500\times\frac{0.674}{0.03277}] [F=11500\times20.568] [F\approx236532]

Answer:

$236532$