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.