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# Summation

## Recommended Posts How do I solve this one?

EDIT: I tried this, I don't think it's right.

Edited by Nebster173
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I've solved it, not even looked it through. But the first step I'd try would be $\sum_{k=1}^{3} \sum_{j=1}^{k} \sum_{i=1}^{j+k} ( i + k -j ) = \sum_{k=1}^{3} \sum_{j=1}^{k} \left[ (j+k)(k-j) + \sum_{i=1}^{j+k} i \right]$.This allows to get rid of the i, as according to a local (and long-dead) mathematician, $\sum_{i=1}^{j+k} i = \frac{(j+k)(j+k+1)}{2}$. I don't quite understand what you tried to do.

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I tried solving it. But I also did it with a calculator and it tells me the result is 81. What am I doing wrong?

EDIT:

I've solved it, not even looked it through. But the first step I'd try would be $\sum_{k=1}^{3} \sum_{j=1}^{k} \sum_{i=1}^{j+k} ( i + k -j ) = \sum_{k=1}^{3} \sum_{j=1}^{k} \left[ (j+k)(k-j) + \sum_{i=1}^{j+k} i \right]$.This allows to get rid of the i, as according to a local (and long-dead) mathematician, $\sum_{i=1}^{j+k} i = \frac{(j+k)(j+k+1)}{2}$. I don't quite understand what you tried to do.

I didn't know you could do that. I don't know anything about the laws of summation.

I figured out what I did wrong. Here's my edited answer. I confirmed it with a calculator and it's correct. Yay! Edited by Nebster173

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