# Need helps with series.

• Aug 29th 2008, 11:58 AM
ceasar_19134
Need helps with series.
"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. • Aug 29th 2008, 12:26 PM Moo Hello, 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.

That's because you didn't consider that a deposit of $500 is made at the beginning of each year Try to reconsider your series (Tongueout) • Aug 29th 2008, 12:35 PM ceasar_19134 The formula I have here is S(n) = a(1)*(1-r^n)/1-r, where n is the number of terms, a(1) is the first term, and r is the common ratio. I'm not sure what to do if that isnt right. • Aug 29th 2008, 02:27 PM Moo Quote: Originally Posted by ceasar_19134 The formula I have here is S(n) = a(1)*(1-r^n)/1-r, where n is the number of terms, a(1) is the first term, and r is the common ratio. I'm not sure what to do if that isnt right. Uuuuh.... That's right (Giggle) Excuse me, I promise I'll concentrate more next time (Headbang) -------------------- I think the problem is as follows : - the interests are added at the end of the year - but you put 500$ at the beginning of the year

Let's show it for 3 years, what happens in the end.
Beginning first year : 500$End first year : 500*1.11 Beginning 2nd year : 500+500*1.11 End 2nd year : 500*1.11+500*1.11² Beginning 3rd year : 500+500*1.11+500*1.11² End 3rd year : 500*1.11+500*1.11²+500*1.11^3 So at the end of the 3rd year, we'll have $500 \cdot \sum_{i=1}^3 1.11^i=500 \cdot 1.11 \cdot \frac{1-1.11^3}{1-1.11}$ (the formula is $\sum_{i=1}^n r^i=\underbrace{r}_{\text{first term}} \cdot \frac{1-r^n}{1-r}$ where n is the number of terms). Which is, in our example $500 \cdot \sum_{i=1}^{25} 1.11^i=500 \cdot {\color{red}1.11} \cdot \frac{1-1.11^{25}}{1-1.11} \approx 63499.3$ Yay ! • Aug 29th 2008, 02:44 PM ceasar_19134 Thank you very much for the help so far, but could you explain where the red 1.11 comes from? I have a few other problems like this one to complete and I want to make sure I'm getting them right. • Aug 30th 2008, 12:10 AM Moo Quote: Originally Posted by ceasar_19134 Thank you very much for the help so far, but could you explain where the red 1.11 comes from? I have a few other problems like this one to complete and I want to make sure I'm getting them right. The formula for a geometric series is : $\text{first term } \times \frac{1-\text{progression}^{\text{number of terms}}}{1-\text{progression}}$ Here, the first term was 1.11, not 1 as usual :p • Aug 30th 2008, 09:12 AM ceasar_19134 This is confusing. Isnt the$500 the first term?

Here is another problem I have, maybe it'll be easier to explain with another.

Suppose you deposit a nickel into your piggy bank on the first day of March and you double the previous day's deposit for every day until March 31. How much will be in the piggy bank at the end of the last day?

A) $53,687,091.15 B)$107,374,182.35
C) $214,748,364.75 D)$21,474,836.47

.05(1-2^30)/(1-2) =$53,687,091.15 Is that correct? Based off of that last one, I would guess this is wrong, but that looks right to me. • Aug 30th 2008, 01:50 PM jonah 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):
$
\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}
$

$
\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}
$

$
\begin{gathered}
\ddot S = 500 \cdot \frac{{({\text{common ratio)(last term)}} - ({\text{first term)}}}}
{{({\text{common ratio)}} - 1}} \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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
$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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}
$

$
\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
$
\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
$
{\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
• Aug 30th 2008, 03:05 PM
jonah
Quote:

Originally Posted by ceasar_19134
This is confusing.

Isnt the $500 the first term? Here is another problem I have, maybe it'll be easier to explain with another. Suppose you deposit a nickel into your piggy bank on the first day of March and you double the previous day's deposit for every day until March 31. How much will be in the piggy bank at the end of the last day? A)$53,687,091.15
B) $107,374,182.35 C)$214,748,364.75
D) $21,474,836.47 .05(1-2^30)/(1-2) =$53,687,091.15

Is that correct? Based off of that last one, I would guess this is wrong, but that looks right to me.

$
\begin{gathered}
.05(2)^0 + .05(2)^1 + .05(2)^2 + \cdot \cdot \cdot + .05(2)^{29} + .05(2)^{30} \hfill \\
= .05 \times \frac{{2^1 \cdot 2^{30} - 1}}
{{2^1 - 1}} \hfill \\
= \ {\text{107,374,182}}{\text{.35}} \hfill \\
\end{gathered}
$