Respuesta :
[tex]\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+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]\left(1+\frac{r}{n}\right)
\\\\
\qquad
[/tex]
[tex]\bf \begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array}\\ pymnt=\textit{periodic payments}\to &2000\\ r=rate\to 8.25\%\to \frac{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}[/tex]
[tex]\bf A=2000\left[ \cfrac{\left( 1+\frac{0.0825}{1} \right)^{1\cdot 35}-1}{\frac{0.0825}{1}} \right]\left(1+\frac{0.0825}{1}\right) \\\\\\ A=2000\left( \cfrac{1.0825^{35}-1}{0.0825} \right)(1.0825)[/tex]
which is very close to 400000.
[tex]\bf \begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array}\\ pymnt=\textit{periodic payments}\to &2000\\ r=rate\to 8.25\%\to \frac{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}[/tex]
[tex]\bf A=2000\left[ \cfrac{\left( 1+\frac{0.0825}{1} \right)^{1\cdot 35}-1}{\frac{0.0825}{1}} \right]\left(1+\frac{0.0825}{1}\right) \\\\\\ A=2000\left( \cfrac{1.0825^{35}-1}{0.0825} \right)(1.0825)[/tex]
which is very close to 400000.
