Compute the below sum:


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.

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

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

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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.

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