# Evaluate Sum - Recurrence Relation

Simple question really that I should be able to solve. I should probably note that the question is the final part of a series of questions involving the difference operator and in the previous part we are asked to find the sum $\sum\limits_{k=1}^n k\cdot k!$ which I managed to solve, but I am not sure whether it's relevant.

Okay, the question is:

Find the sum $\sum \limits_{k=1}^n k(k+1)2^k$.

My initial thoughts are that perhaps we are supposed to use some cleverness with recurrence relations, possibly using the Repetoire Method. The problem with this is that I don't know how to apply this method with the $2^k$ term in there.

Any help would be greatly appreciated.

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$$\sum_{k=1}^nk(k+1)q^k=\frac\partial{\partial q}\sum_{k=1}^nkq^{k+1}=\frac\partial{\partial q}q^2\sum_{k=1}^nkq^{k-1}=\frac\partial{\partial q}q^2\frac\partial{\partial q}\sum_{k=1}^nq^k\;.$$
why not just $\frac{\partial^2}{\partial q^2}q \sum_{k=0}^{n}q^k$? –  Alex Jun 4 '13 at 17:41