Simplify $\sum_{i=1}^{n}\sum_{j=i+1}^{n} (i+j)$ $$\sum_{i=1}^n \sum_{j=i+1}^n (i+j)$$
I'm pretty sure it's somewhat simple to solve but I can't get it done.
I know that the given sum is equal to $\frac{n}{2} (n^2-1)$.
I've been on this for 3h+ now and I'm really hoping someone can give me a hint :)

By double counting with $p=i+j$

we have
$$\sum_{i=0}^n \sum_{j=i+1}^n (i+j)=\frac12\left(\sum_{p=1}^{n} p(p-1)+\sum_{p=n+1}^{2n+1} (2n-p+2)(p-1)\right)-\sum_{p=1}^{n}p=$$
$$=\frac12\left(\sum_{p=1}^{n} (p^2-3p)+\sum_{p=1}^{n+1} (n-p+2)(n+p-1)\right)=$$
$$=\frac12\left(\sum_{p=1}^{n} (n^2+n-2)+2n\right)=\frac12n(n^2+n-2)+n=\frac12n^2(n+1)$$
The answer you are referring to is for the sum starting form $i=1$ indeed
$$\frac12n^2(n+1)-\frac12 n(n+1)=\frac12n(n^2-1)$$
