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.
$$ \sum_{k=1}^nk(k+1)q^k=\frac\partial{\partial q}\sum_{k=1}^nkq^{k+1}=\frac\partial{\partial q}q^2\sum_{k=1}^nkq^{k-1}=\frac\partial{\partial q}q^2\frac\partial{\partial q}\sum_{k=1}^nq^k\;. $$