Assume that $50 is deposited at the end of each period in an account in which interest is compounded at the frequency k = 12, at an annual rate of 9%.Find the accumulated amount after 3 years.Round to the
Nearest cent.
A)$2,033.84
B)$2,052.39
C)$2,057.64
D)$2,075.84
E)$2,047.14

Question

Quizplus · Accepted Answer

We can use the formula for compound interest with regular deposits:

$A=P\left(1+\frac{r}{k}\right)^{kt}\left(\frac{(1+r/k)^{kt}-1}{r/k}\right)+PMT\frac{(1+r/k)^{kt}-1}{r/k}$

where $P$ is the initial balance (which is 0 in this case), $r$ is the annual interest rate (9%), $k$ is the compounding frequency per year (12), $t$ is time in years (3), and $PMT$ is the regular deposit amount ($50).

Plugging in these values, we get:

$A=0\left(1+\frac{0.09}{12}\right)^{12\cdot 3}\left(\frac{(1+0.09/12)^{12\cdot 3}-1}{0.09/12}\right)+50\frac{(1+0.09/12)^{12\cdot 3}-1}{0.09/12}$

Simplifying, we get:

$A=50\cdot\frac{(1+0.09/12)^{36}-1}{0.09/12}$

Using a calculator, we get $A\approx 2057.64,$ so the answer is (C) $2,057.64.$

Assume That $50 Is Deposited at the End of Each