TITLE: Check computation of: $\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}ij$
QUESTION [5 upvotes]: Compute the below sum:
$\sum_{i=1}^{n}\sum_{j=1}^{n}ij$
My working:
$\sum_{i=1}^{n}\sum_{j=1}^{n}ij = \sum_{i=1}^{n}i\frac{n(n+1)}{2}$
Now since $\frac{n(n+2)}{2}$ is just a constant we can take it out of the sum
$\sum_{i=1}^{n}i\frac{n(n+1)}{2} = \frac{n(n+1)}{2}\sum_{i=1}^{n}i$
$\sum_{i=1}^{n}i\frac{n(n+1)}{2} = \frac{n(n+1)}{2}\times\frac{n(n+1)}{2}$
So we get:
$\sum_{i=1}^{n}\sum_{j=1}^{n}ij = (\frac{n(n+1)}{2})^2$
I am not sure if this is correct, or if I am using the properties of Series correctly.

REPLY [4 votes]: another way:
$\sum_{i=1}^{n}\sum_{j=1}^{n}ij=\sum_{i=1}^{n}i * \sum_{j=1}^{n}j=\frac{n(n+1)}{2} * \frac{n(n+1)}{2}$

REPLY [2 votes]: Your proof is correct. 
Since $n(1+2+3+\ldots+n-1)n+n(1+2+3+\ldots+n-1)+n^2=n^3$ using induction it is not hard to show that $\sum_{i=1}^n\sum_{j=1}^nij=\sum_{k=1}^nk^3$.
Therefore you also have $\sum_{k=1}^nk^3=\left(\dfrac{n(n+1)}{2}\right)^2$. Of course if you know the last identity you can use it for a different proof of your result.