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\[\sum_{i=1}^{n}i^2r^{i-1}\]

What have you tried so far?

actually it is identical xD

The second method uses some kind of convolution of two interleaving sequences

we can simply plugin the previous result for the sum in middle, awesome!

I know I made a small error in it

that was a typo in the exponent, its okay it is a beautiful solution !

Hmm, I still shouldn't have that n term, they should be m most likely

what is r here?

I think r could be any real number

The whole point though is to get rid of that outer sum. And you shouldn't have any \(n\) terms.

"n" is the index, nth partial sum right

oh, right, I forgot

Everything goes to hell when you let \(r=1\) though.

\[
f(m, 1) = \sum_{k=0}^nk^m
\]

Haha lets just say \(r\ne 1\) then

But \(r=1\) is an interesting series.

Is there a formula for that series?

I think \(f(-1,1)\) is known not to have an elementary function.

Sorry it is not a formula, just a limit which gives the asymptotic sum

\[\lim\limits_{n\to\infty}\dfrac{f(m,1)}{n^{m+1}} = \frac{1}{m+1}\]

for r=1 sum if n(n+1)(2n+1)/6 you can prove using induction

http://math.stackexchange.com/questions/936924/limit-of-the-sequence-frac1k2k-nknk1

like in sum of squares of first n

you mean explicit formula

i don't know what it is called

i don't know math's english terminology very well :)

yeah the original question is a special case, \(m=2\)

Oh I thought he just did the derivatives of the geometric series approach.

He did it in two ways..

and do we know that there is a formula for this?

ok thanks

generating functions might also work i think
post the work @ikram002p