ABCD1234pop 1 Posted September 15, 2012 Share Posted September 15, 2012 (edited) How do I solve this one? EDIT: I tried this, I don't think it's right. Edited September 15, 2012 by Nebster173 Link to post Share on other sites

timo 583 Posted September 15, 2012 Share Posted September 15, 2012 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. Link to post Share on other sites

ABCD1234pop 1 Posted September 15, 2012 Author Share Posted September 15, 2012 (edited) 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 [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. I confirmed it with a calculator and it's correct. Yay! Edited September 15, 2012 by Nebster173 Link to post Share on other sites

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