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Few days ago, I found this equation:

$ \sum_{i=1}^n \sum_{j>i} \frac{1}{2} = {n \choose 2} \frac{1}{2} $

I didn't manage to prove it. Does anyone of you know how to prove it?

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$\sum_{i=1}^n \sum_{j>i} \frac{1}{2}=\sum_{i=1}^n (n-i) \frac{1}{2} = \big(n^2-\frac{n(n+1)}{2}\big)\frac{1}{2} = {n \choose 2} \frac{1}{2} $

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  • $\begingroup$ how did you get from (1) to (2)? [1] $(n-1) + (n-2) + ... + (n-n)$ [2] $(n^2 - \frac{n(n+1)}{2})$ $\endgroup$ – user1384636 Oct 17 '14 at 19:25
  • $\begingroup$ now it's written properly. $\endgroup$ – user1384636 Oct 17 '14 at 19:29
  • $\begingroup$ @user1384636 group $n$ together, you get $n^2$, and $(1+2+\cdots +n)=\frac{(n+1)n}{2}$ $\endgroup$ – John Oct 17 '14 at 19:36
  • $\begingroup$ yes! I'm starting to lose my mind $\endgroup$ – user1384636 Oct 17 '14 at 19:59

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