future value on annuity due

Quote:

Originally Posted by

**ceasar_19134** "Suppose a deposit of $500 is made at the begining of each year for 25 years into an account that pays 11% compounded annually. What is the amount of this annunity at the end of the 25th year?"

I think the formula I should be using is one for a geometric series so I tried it:

500(1-1.11^25)/(1-1.11) = 57206.65

However, that doesnt match with any of my answers. Please explain to me where I am going wrong.

Another way of looking at your problem is by considering the following procedure (the standard procedure I’m used to):

$\displaystyle

\begin{gathered}

\ddot S = 500\left( {1 + \tfrac{{.11}}

{1}} \right)^1 + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^2 \hfill \\

+ \cdot \cdot \cdot + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1 - 2} + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1 - 1} + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500\left[ \begin{gathered}

\left( {1 + \tfrac{{.11}}

{1}} \right)^1 + \left( {1 + \tfrac{{.11}}

{1}} \right)^2 + \cdot \cdot \cdot \hfill \\

+ \left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1 - 2} + \left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1 - 1} + \left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} \hfill \\

\end{gathered} \right] \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{({\text{common ratio)(last term)}} - ({\text{first term)}}}}

{{({\text{common ratio)}} - 1}} \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)^1 \left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} - \left( {1 + \tfrac{{.11}}

{1}} \right)^1 }}

{{\left( {1 + \tfrac{{.11}}

{1}} \right)^1 - 1}} \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} \cdot \left( {1 + \tfrac{{.11}}

{1}} \right)^1 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = \frac{{500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} - 500 + \tfrac{{.11}}

{1} \cdot 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} }}

{{\tfrac{{.11}}

{1}}} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} + \tfrac{{.11}}

{1}\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\ddot S = 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1 + 1} - 1}}

{{\tfrac{{.11}}

{1}}} - 500

$

Still another way of looking at it is:

Future value of an annuity due (beginning of interval deposits)

= future value of 1st deposit at the end of 25 years + future value of the next 24 deposits at the end of 25 years

or

$\displaystyle

\begin{gathered}

\ddot S = 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} \hfill \\

+ \left[ {\underbrace {500 + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{1 \times 1} + \cdot \cdot \cdot + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{22 \times 1} + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{23 \times 1} }_{{\text{24 terms}}}} \right]\left( {1 + \tfrac{{.11}}

{1}} \right) \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} \hfill \\

+ 500\left[ {1 + \left( {1 + \tfrac{{.11}}

{1}} \right)^{1 \times 1} + \cdot \cdot \cdot + \left( {1 + \tfrac{{.11}}

{1}} \right)^{22 \times 1} + \left( {1 + \tfrac{{.11}}

{1}} \right)^{23 \times 1} } \right]\left( {1 + \tfrac{{.11}}

{1}} \right) \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} \hfill \\

+ 500 \cdot \frac{{({\text{common ratio)(last term)}} - ({\text{first term)}}}}

{{({\text{common ratio)}} - 1}} \cdot \left( {1 + \tfrac{{.11}}

{1}} \right) \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} \hfill \\

+ 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)^{1 \times 1} \left( {1 + \tfrac{{.11}}

{1}} \right)^{23 \times 1} - 1}}

{{\left( {1 + \tfrac{{.11}}

{1}} \right)^{1 \times 1} - 1}} \cdot \left( {1 + \tfrac{{.11}}

{1}} \right) \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} \hfill \\

+ 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)^{24 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} \cdot \left( {1 + \tfrac{{.11}}

{1}} \right) \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} \hfill \\

+ 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)^{24 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{24 \times 1} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = \frac{{\tfrac{{.11}}

{1} \cdot 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} + 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{24 \times 1} - 500 + \tfrac{{.11}}

{1} \cdot 500\left( {1 + \tfrac{{.11}}

{1}} \right)^{24 \times 1} }}

{{\tfrac{{.11}}

{1}}} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{\tfrac{{.11}}

{1}\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} + \left( {1 + \tfrac{{.11}}

{1}} \right)^{24 \times 1} + \tfrac{{.11}}

{1}\left( {1 + \tfrac{{.11}}

{1}} \right)^{24 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{\tfrac{{.11}}

{1}\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} + \left( {1 + \tfrac{{.11}}

{1}} \right)\left( {1 + \tfrac{{.11}}

{1}} \right)^{24 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{\tfrac{{.11}}

{1}\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} + \left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

\ddot S = 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1} - 1}}

{{\tfrac{{.11}}

{1}}} - 500 \hfill \\

\Leftrightarrow \hfill \\

\end{gathered}

$

$\displaystyle

\ddot S = 500 \cdot \frac{{\left( {1 + \tfrac{{.11}}

{1}} \right)^{25 \times 1 + 1} - 1}}

{{\tfrac{{.11}}

{1}}} - 500

$

I merely aligned both cases towards the so-called standard formula for the future value of an annuity due (beginning of interval deposits) as given by

$\displaystyle

\ddot S = R \cdot s_{\left. {\overline {\,

{n + 1} \,}}\! \right| i} - R = R \cdot \frac{{\left( {1 + i} \right)^{n + 1} - 1}}

{i} - R = R \cdot \frac{{\left( {1 + \tfrac{j}

{m}} \right)^{tm + 1} - 1}}

{{\tfrac{j}

{m}}} - R

$

where

$\displaystyle

{\ddot S}

$ = future value of an annuity due

* R* = periodic deposits ($500 annually in this case)

* j* = nominal interest rate (.11 as in 11%)

* m* = interest period (1 as in annually)

* t* = time from 1st deposit to end of one period after last deposit (25 years)

As you can see, this is but another way of arriving at Moo’s result of approximately $63,499.39