# 022 Sample Final A, Problem 10

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${\displaystyle A\,=\,P\left(1+{\frac {r}{n}}\right)^{nt},}$
where ${\displaystyle A}$ is the value of the account, ${\displaystyle P}$ is the principal (original amount invested), ${\displaystyle r}$ is the annual rate and ${\displaystyle n}$ is the number of compoundings per year. The value of ${\displaystyle n}$ is ${\displaystyle 365}$ for compounding daily, ${\displaystyle 52}$ for compounding weekly, and ${\displaystyle 12}$ for compounding monthly. As a result, the exponent ${\displaystyle nt}$, where ${\displaystyle t}$ is the time in years, is the number of compounding periods where we actually earn interest. Similarly, ${\displaystyle r/n}$ is the rate per compounding period (the annual rate divided by the number of compoundings per year).
For example, if we compound monthly for ${\displaystyle 7}$ years at a ${\displaystyle 6\%}$ rate, we would compound ${\displaystyle nt\,=\,12\cdot 7\,=\,84}$ times, once per month, at a rate of ${\displaystyle 0.06/12\,=\,0.005}$ per monthly period. Notice that we always use the decimal version for interest rates when using these equations.
We are given all the pieces required. We begin with ${\displaystyle \2100}$ of principal, and compound quarterly, or ${\displaystyle 4}$ times per year. Using the formula in 'Foundations', the equation for the account value after 8 years is
${\displaystyle A\,=\,P\left(1+{\frac {r}{n}}\right)^{nt}\,=\,2100\left(1+{\frac {0.06}{4}}\right)^{4\cdot 8}\,=\,2100(1.015)^{32}.}$
${\displaystyle A\,=\,2100(1.015)^{32}.}$