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Suppose $$X = 1.05^{35}v+1.05^{34}v^{2} + \cdots + 1.05v^{35}$$ where $v = 1/1.05$. Then we have $$X = 1.05^{35}v(1+ 0.952v+ \cdots + 0.952^{34}v^{34})$$

So the sum of this would be $$ X = 1.05^{35}v \left[\frac{1-\left(\frac{1+952v}{1.05} \right)^{35}}{1-\frac{1+.952v}{1.05}} \right]$$

Is that right?

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Except that $1/1.05\ne .952$. (this will be important when taking a big power) – David Mitra Dec 4 '11 at 20:51
So, to clarify, the sum in question is $X=\sum_{i=1}^{35} 1.05^{36-i}\left(\frac{1}{1.05}\right)^{i}$ ? – Drew Christianson Dec 4 '11 at 21:50
Additionally, what is the context of this problem? It looks closest to finding the PV of a growing annunity, but its not exactly the same. – Drew Christianson Dec 4 '11 at 23:01

The formula for the sum of a finite geometric series is $\sum_{k=0}^n ar^k = \frac{a(1-r^{n+1})}{1-r}$. You correctly factored out $a=1.05^{35}v$, so identifying $r=v/1.05$, we have, $$\begin{align} X &= 1.05^{35}v+1.05^{34}v^{2} + \cdots + 1.05v^{35} \\ &= 1.05^{35}v\left(1+\frac{v}{1.05}+\frac{v^2}{1.05^2}+\cdots+\frac{v^{34}}{1.05^{34}}\right) \\ &= 1.05^{35}v \left(\sum_{k=0}^{34} \left(\frac{v}{1.05}\right)^k\right) \\ &= \frac{1.05^{35}v\left(1-\left(\frac{v}{1.05}\right)^{35}\right)}{1-\frac{v}{1.05}}. \end{align}$$

Your identification of $r$ is not quite correct.

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I just rounded $r$. – james Dec 4 '11 at 20:57
If you didn't round, your choice would be $r=\frac{1+(1/1.05)}{1.05}v$. This is not the same as $r=\frac{v}{1.05}$. – dls Dec 4 '11 at 21:00
Your $r$ and my $r$ are in different contexts. I am using the following formula: $S = (\text{first term}) \frac{[1-(\text{ratio})^{\text{no. of terms}}]}{1-\text{ratio}}$ where $\text{ratio} = \frac{1+r}{1+i}$. So my $r$ is $1/1.05^{2}$. – james Dec 4 '11 at 21:15
I am using $r$ as the ratio of terms in the series, which is simply $v/1.05$. The formula you posted for the ratio seems to be related to interest rates, which wasn't explicitly mentioned in the original question. – dls Dec 4 '11 at 21:39
@james $r=\frac{v}{1.05}$ in dls's answer is $\frac{1}{1.05^2}$ as well. – Drew Christianson Dec 4 '11 at 22:52

So I worked through the entire problem, and then realized I'd done exactly what dls already had posted. However, I arrived at the answer in a slightly different manner (eliminating the v) that felt slightly more intuitive to me. So:

If I'm interpreting your sequence correctly, you want the sum: $$x=\sum_{i=1}^{35}1.05^{36-i}*\left(\frac{1}{1.05}\right)^i = 1.05^{35}\left(\frac{1}{1.05}\right)^{1}+1.05^{34}\left(\frac{1}{1.05}\right)^{2} + \cdots + 1.05^{1}\left(\frac{1}{1.05}\right)^{35}$$

Then we factor $1.05^{35}\left(\frac{1}{1.05}\right)^{}$, giving:

$$1.05^{35}\left(\frac{1}{1.05}\right)^{}\left[1 + \left(\frac{1}{1.05}\right)^{2}+\left(\frac{1}{1.05}\right)^{4}+\cdots+\left(\frac{1}{1.05}\right)^{68}\right]$$

$$ = 1.05^{35}\left(\frac{1}{1.05}\right)^{}\left[\left(\left(\frac{1}{1.05}\right)^2\right)^{0}+\left(\left(\frac{1}{1.05}\right)^2\right)^{1}+\left(\left(\frac{1}{1.05}\right)^2\right)^{2}+\cdots+\left(\left(\frac{1}{1.05}\right)^2\right)^{34}\right] $$

$$=1.05^{35}\left(\frac{1}{1.05}\right)^{} \sum_{i=0}^{34}\left(\left(\frac{1}{1.05}\right)^2\right)^{i}$$

Which is just a geometric series with $r=\left(\frac{1}{1.05}\right)^2$. Thus, X is given by the formula $\sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r}$. Filling in the values from above:

$$X=1.05^{35}\left(\frac{1}{1.05}\right)^{}\frac{1-\left[\left(\frac{1}{1.05}\right)^2\right]^{35}}{1-\left(\frac{1}{1.05}\right)^2} = 54.6484...$$

Finally, I verified the final value numerically in maple and everything checks out.

On an aside regarding david's comment on rounding: I think this derivation lets you see why rounding $\frac{1}{1.05}$ to .952 causes problems. The final term is a 68th power, so that small error can grow. Numerically, $.952^{68} = .0352624\ldots$ while $\left(\frac{1}{1.05}\right)^{68} = .036234\ldots$ A thousandth might not seem like much, but over 35 terms, you could be off by a third of a one percent - significant for large financial calculations. Moreover, if this was monthly or daily interest (and thus you were taking powers to the hundreds or thousands) those errors could really add up.

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Hence the frequent admonition: round only at the end! Keep as many figures as you can in intermediate calculations; it's not always easy to tell which parts of your long calculation will be a source of grief. – J. M. Dec 5 '11 at 0:32

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