Let $\{S_k\}$ be a sequence of numbers. Then by reversing the sum we have the following, $$A=\sum_{k=0}^n k\cdot S_k \cdot S_{n-k} = \sum_{k=0}^n (n-k) \cdot S_{n-k} \cdot S_{k} .$$ Thus adding these sums together gives us, $$A+A=\sum_{k=0}^n n\cdot S_{k}\cdot S_{n-k},$$ which means that $$A=\frac{n}{2} \sum_{k=0}^n S_k \cdot S_{n-k}.$$ Thus, we have taken the $k$ that was dependent in the sum and removed it as an independent $\frac{n}{2}$. Is there a similar trick for $$B=\sum_{k=0}^n k^2 \cdot S_k\cdot S_{n-k}?$$ That is, how do I get ride of that $k^2$? Any help would be greatly appreciated, including a resourceful book on sums and series manipulations.
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This is pretty easy while $S_k=S(k)$ and $S(k)$ is differentiable. But there is no formula in common case, i think, because each summand has the different coefficient in this sum and the summands are independent. But maybe... – gukoff Jun 08 '13 at 11:13
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How would you re-write this with differentiation? – Bobby Ocean Jun 08 '13 at 20:07
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This is pretty easy, it is the standard trick. When you want to find $\sum\limits_{i=0}^{n}k\cdot f(k)$, you use the fact that $\sum\limits_{i=0}^{n}k\cdot f(k)\cdot x^k = x\left(\sum\limits_{i=0}^{n}f(k)\cdot x^k\right)'$. (2 times for $k^2$). I've mistaken, $f$ shouldn't be differentiable, it should be integrable and you should know $\sum\limits_{i=0}^{n}f(k)\cdot x^k$. It is often easy to find, in fact. The last thing you need to do is to set $x$ to $1$. – gukoff Jun 09 '13 at 02:08