Suppose that Helen deposits $\$2000$ at the end of each month into an account that pays $6\%$ interest per year compounded monthly. The amount of interest she has accumulated after $n$ months is given by the sequence.
$\displaystyle A_n = 2000 \left( 1 + \frac{0.024}{12} \right)$
a.) Find the first six terms of the sequence.
b.) Find the amount in the account after 3 years.
a.)
$
\begin{equation}
\begin{aligned}
i && I_1 &= 100 \left( \frac{1.005^{(1)} - 1}{0.005} - (1) \right) = 0\\
\\
ii&& I_2 &= 100 \left( \frac{1.005^{(2)} - 1}{0.005} - (2) \right) = 0.5\\
\\
iii&& I_3 &= 100 \left( \frac{1.005^{(3)} - 1}{0.005} - (3) \right) = 1.50\\
\\
iv&& I_4 &= 100 \left( \frac{1.005^{(4)} - 1}{0.005} - (4) \right) = 3.01\\
\\
v&& I_5 &= 100 \left( \frac{1.005^{(5)} - 1}{0.005} - (5) \right) = 5.03\\
\\
vi&& I_6 &= 100 \left( \frac{1.005^{(6)} - 1}{0.005} - (6) \right) = 7.55\\
\end{aligned}
\end{equation}
$
b.) If $\displaystyle n = 5 \text{years} \times \frac{\text{12 months}}{\text{1 year}} = 60 \text{months}$, then
$\displaystyle I_{60} = 100 \left( \frac{1.005^{(60)} - 1}{0.005} - (60) \right) = \$977$
It shows that the interest has accumulated $\$977$ after $5$ years.
Monday, October 29, 2018
College Algebra, Chapter 9, 9.1, Section 9.1, Problem 72
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