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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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$$\sum_{k=0}^{n-1}k\cdot ar^k=ar\sum_{k=1}^{n-1}kr^{k-1}=ar\frac{d}{dr}\left(\sum_{k=1}^{n-1}r^k\right)= \frac{ar(r-r^n)}{1-r}$$ It’s been derived many times here, but I don’t immediately find an example – Brian M. Scott Nov 5 '12 at 21:27
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. – Mark Bennet Nov 5 '12 at 21:27

\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 $$\frac{ar(nr-1)(r^n-1)}{(r-1)^2}.$$ – Cameron Buie Nov 5 '12 at 22:25

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