5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.