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

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I tried solving it.

 

IMG_20120915_230928.jpg

 

 

 

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 [math] \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][/math].This allows to get rid of the i, as according to a local (and long-dead) mathematician, [math] \sum_{i=1}^{j+k} i = \frac{(j+k)(j+k+1)}{2}[/math]. 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.

 

IMG_20120915_232603.jpg

 

I confirmed it with a calculator and it's correct. Yay! :D

Edited by Nebster173
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