Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$ \sum_{k=0}^{n} k\cdot ar^k?$$

share|improve this question
1  
$$\sum_{k=0}^{n-1}k\cdot ar^k=ar\sum_{k=1}^{n-1}kr^{k-1}=ar\frac{d}{dr}\left(\sum_{k=1}^{n-1}r^k\right)= \frac{ar(r-r^n)}{1-r}$$ 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
add comment

1 Answer

\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}

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.