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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.

GraphicsMuncher
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Benjamin Biggs
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1 Answers1

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

joriki
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  • why not just $\frac{\partial^2}{\partial q^2}q \sum_{k=0}^{n}q^k$? – Alex Jun 04 '13 at 17:41
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    @Alex: You're right, that's more elegant -- but I'm not sure it's actually easier to calculate, since you're taking the derivative of a product twice whereas my first derivative doesn't create quite as many terms. – joriki Jun 04 '13 at 18:14
  • Thanks very much for your answers. The only thing is that I don't think I'm expected to know (at least for this exam) how to evaluate sums by taking derivatives. Do you know of any other way of solving the problem without the need for derivatives? – Benjamin Biggs Jun 04 '13 at 18:53