# Proving summation identities [duplicate]

How would one go about proving the following identities?

$$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i}{z_i-z_j} = \frac{n(n-1)}{2}$$ $$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^2}{z_i-z_j} = (n-1)\sum_{i=1}^n z_i$$ $$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^3}{z_i-z_j} = (n-1)\sum_{i=j}^n z_i^2+\sum_{i<j}^n z_i z_j$$ $$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^4}{z_i-z_j} = (n-1)\sum_{i=j}^n z_i^3+\sum_{i<j}^n z_i z_j(z_i+z_j)$$ $$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^5}{z_i-z_j} = (n-1)\sum_{i=j}^n z_i^4+\sum_{i<j}^n z_i z_j(z_i^2+z_i z_j +z_j^2)$$

I see the obvious pattern here. The problem is that the algebra involving the summation is giving me some difficulty.

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## marked as duplicate by Peter Taylor, TMM, Davide Giraudo, Ayman Hourieh, JonathanJan 12 '14 at 17:35

What is $\large\left\{z_{i}\right\}$ ?. – Felix Marin Jan 12 '14 at 2:33

$\sum_{i=1}^{n}\sum_{i \neq j}^n \frac{z_i}{z_i-z_j}$ $=\sum_{i=1}^{n}\sum_{i \neq j}^n \left(1+\frac{z_j}{z_i-z_j} \right)=n(n-1)-\sum_{i=1}^{n}\sum_{i \neq j}^n \frac{z_j}{z_i-z_j}.$
You can now use the same strategy to split $\frac{z_i^2}{z_i-z_j}=\frac{(z_i^2-z_j^2)+z_j^2}{z_i-z_j}=(z_i+z_j)+\frac{z_j^2}{z_i-z_j}$