Respuesta :
Fv=200×(((1+0.0295÷12)^(41
×12)−1)÷(0.0295÷12))×(1+0.0295÷12)
Fv=191,398.48
he's 24 right now, and retiring at 65 means 41 years later.
[tex]\bf ~~~~~~~~~~~~\textit{Future Value of an annuity due}\\ ~~~~~~~~~~~~(\textit{payments at the beginning of the period}) \\\\ A=pmt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]\left(1+\frac{r}{n}\right) \\\\ [/tex]
[tex]\bf \qquad \begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array} \\ pmt=\textit{periodic payments}\to &200\\ r=rate\to 2.95\%\to \frac{2.95}{100}\to &0.0295\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{they're done monthly, thus} \end{array}\to &12\\ t=years\to &41 \end{cases}[/tex]
[tex]\bf A=200\left[ \cfrac{\left( 1+\frac{0.0295}{12} \right)^{12\cdot 41}-1}{\frac{0.0295}{12}} \right]\left(1+\frac{0.0295}{12}\right) \\\\\\ A=200\left[ \cfrac{\left( 1 +\frac{59}{24000}\right)^{492}-1}{\frac{59}{24000}}\right]\left(1+\frac{59}{24000}\right) \\\\\\ A=200\left[ \cfrac{\left( \frac{24059}{24000}\right)^{492}-1}{\frac{59}{24000}}\right]\left(\frac{24059}{24000}\right)\\\\\\A\approx 191398.48860411668565454023[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{Future Value of an annuity due}\\ ~~~~~~~~~~~~(\textit{payments at the beginning of the period}) \\\\ A=pmt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]\left(1+\frac{r}{n}\right) \\\\ [/tex]
[tex]\bf \qquad \begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array} \\ pmt=\textit{periodic payments}\to &200\\ r=rate\to 2.95\%\to \frac{2.95}{100}\to &0.0295\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{they're done monthly, thus} \end{array}\to &12\\ t=years\to &41 \end{cases}[/tex]
[tex]\bf A=200\left[ \cfrac{\left( 1+\frac{0.0295}{12} \right)^{12\cdot 41}-1}{\frac{0.0295}{12}} \right]\left(1+\frac{0.0295}{12}\right) \\\\\\ A=200\left[ \cfrac{\left( 1 +\frac{59}{24000}\right)^{492}-1}{\frac{59}{24000}}\right]\left(1+\frac{59}{24000}\right) \\\\\\ A=200\left[ \cfrac{\left( \frac{24059}{24000}\right)^{492}-1}{\frac{59}{24000}}\right]\left(\frac{24059}{24000}\right)\\\\\\A\approx 191398.48860411668565454023[/tex]