Check computation of: $\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}ij$

Question

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.

You are correct. One way to check is to do a few examples - n=1,2,3. — Thomas Andrews, Apr 29 '13 at 12:53
Generally correct; in your first line I would pull the $i$ out of the inner sum to justify replacing it with $\frac{n(n+1)}{2}$. — vadim123, Apr 29 '13 at 12:53
incidentally $\sum_{i=1}^n\sum_{j=1}^n ij = (\sum_{i=1}^ni)(\sum_{j=1}^nj)$ — Gautam Shenoy, Apr 29 '13 at 12:59

score 4 · Accepted Answer · answered Apr 29 '13 at 13:06

4

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

answered Apr 29 '13 at 13:06

Somaye

2,730

score 2 · Answer 2 · answered Apr 29 '13 at 13:07

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.

Check computation of: $\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}ij$

2 Answers2