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I need a little help in summing the the following series:

$ 1+2v^4+3v^8+4v^{12}+\ldots + 20v^{76}?$


Is there a closed formula for summing $ \sum_{k=0}^{n} k\cdot ar^k?$

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    There are various approaches, but you could try here setting $x=v^4$. Then sum $x+x^2+x^3+ \dots$ and then differentiate both sides.2012-11-05

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\begin{align} S & = \sum_{k=1}^n akr^k = ar \sum_{k=1}^{n} k r^{k-1} = ar \dfrac{d}{dr} \left( \sum_{k=1}^{n} r^{k}\right)\\ & = ar \dfrac{d}{dr} \left( \dfrac{r(r^n-1)}{r-1}\right) = ar \left(\dfrac{nr^{n+1} - (n+1)r^n + 1}{(r-1)^2} \right) \end{align}

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    If you felt like it, you could factor that last bit to $\frac{ar(nr-1)(r^n-1)}{(r-1)^2}.$2012-11-05