022 Sample Final A, Problem 10

Set up the formula to find the amount of money one would have at the end of 8 years if she invests $2100 in an account paying 6% annual interest, compounded quarterly.

Foundations:
The primary purpose of this problem is to demonstrate that you understand compounding on an interval of time. When we compound on an interval, say monthly, the value in the account only changes at the end of each interval. In other words, there is no interest accrued for a week or a day. As a result, we use the formula
$A\,=\,P\left(1+{\frac {r}{n}}\right)^{nt},$
where $A$ is the value of the account, $P$ is the principal (original amount invested), $r$ is the annual rate and $n$ is the number of compoundings per year. The value of $n$ is $365$ for compounding daily, $52$ for compounding weekly, and $12$ for compounding monthly. As a result, the exponent $nt$, where $t$ is the time in years, is the number of compounding periods where we actually earn interest. Similarly, $r/n$ is the rate per compounding period (the annual rate divided by the number of compoundings per year).
For example, if we compound monthly for $7$ years at a $6\%$ rate, we would compound $nt\,=\,12\cdot 7\,=\,84$ times, once per month, at a rate of $0.06/12\,=\,0.005$ per monthly period. Notice that we always use the decimal version for interest rates when using these equations.

Solution:
We are given all the pieces required. We begin with $\$2100$ of principal, and compound quarterly, or $4$ times per year. Using the formula in 'Foundations', the equation for the account value after 8 years is
$A\,=\,P\left(1+{\frac {r}{n}}\right)^{nt}\,=\,2100\left(1+{\frac {0.06}{4}}\right)^{4\cdot 8}\,=\,2100(1.015)^{32}.$

Final Answer:
$A\,=\,2100(1.015)^{32}.$

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