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 :)
 A: Here is a more visual approach: Fill a square $n\times n$ grid with the values of $i+j$:
$$
\begin{matrix}
1+1&1+2&1+3&\cdots&1+n\\
2+1&2+2&2+3&\cdots&2+n\\
3+1&3+2&3+3&\cdots&3+n\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
n+1&n+2&n+3&\cdots&n+n
\end{matrix}
$$
We are after the sum of all the elements above the main diagonal (the diagonal that goes from $1+1$ to $n+n$). Note that this is equal to the sum of all the elements below the main diagonal. So if we can find the sum of the entire table, and the sum of the main diagonal, then a simple subtraction followed by a division by 2 will get us there.
The sum of the main diagonal is $n(n+1)$.
For the sum of the entire table, collect each $j$ column-by column, and all the $i$'s row-by-row. The $j$'s in the first column sum to $n$. The $j$'s in the second column sum to $2n$, and so on. And similarly for the $i$'s. Thus the sum of all the elements in the table is
$$
\underbrace{n+2n+\cdots+n^2}_{\text{the }j\text{'s}}+\underbrace{n+2n+\cdots+n^2}_{\text{the }i\text{'s}}=2n(1+2+\cdots+n)\\
=n^2(n+1)
$$
Our final answer is therefore
$$
\frac{n^2(n+1)-n(n+1)}2=\frac12n(n^2-1)
$$
A: Here is to help you to start:
\begin{align}
\sum_{i=1}^n \sum_{j=i+1}^n (i+j) &= \left( \sum_{i=1}^n \sum_{j=i+1}^n i\right)+\left( \sum_{i=1}^n \sum_{j=i+1}^n j\right) \\
&= \left( \sum_{i=1}^n i (n-(i+1)+1)\right)+\left( \sum_{i=1}^n \left(\sum_{j=1}^n j -\sum_{j=1}^i j \right)\right) \\
\end{align}
Try to simplify it. Formulas to evaluate $\sum i^2$ is useful as well.
A: 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
$$\frac12n^2(n+1)-\frac12 n(n+1)=\frac12n(n^2-1)$$
