# Binomial Coefficient as Sum of a Sum

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?

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

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