Question

John decides to invest $2,000 per year and an account that he opened and expects to earn 8.25%. If he keeps this up for 35 years, how much would he end up with in his account?

Answer 1

`\bf \qquad \qquad \textit{Future Value of an annuity due}\\ \left. \qquad \right.(\textit{payments at the beginning of the period}) \\\\ A=pymnt\left[ \cfrac{\left( 1+(r)/(n) \right)^(nt)-1}{(r)/(n)} \right]\left(1+(r)/(n)\right) \\\\ \qquad`


`\bf \begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array}\\ pymnt=\textit{periodic payments}\to &2000\\ r=rate\to 8.25\%\to (8.25)/(100)\to &0.0825\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &35 \end{cases}`


`\bf A=2000\left[ \cfrac{\left( 1+(0.0825)/(1) \right)^(1\cdot 35)-1}{(0.0825)/(1)} \right]\left(1+(0.0825)/(1)\right) \\\\\\ A=2000\left( \cfrac{1.0825^(35)-1}{0.0825} \right)(1.0825)`

which is very close to 400000.