Difference between revisions of "Talk:TADM2E 2.36"

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(Created page with " == Proposed solution (with steps) == <math>\sum\limits_{i=1}^{n/2} \sum\limits_{j=i}^{n-i} \sum\limits_{k=1}^{j} 1</math> The innermost summation is just j since the distan...")
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Revision as of 00:45, 13 December 2014

Proposed solution (with steps)

$ \sum\limits_{i=1}^{n/2} \sum\limits_{j=i}^{n-i} \sum\limits_{k=1}^{j} 1 $

The innermost summation is just j since the distance in the progression is 1 and the coefficient is also 1.

$ \sum\limits_{i=1}^{n/2} \sum\limits_{j=i}^{n-i} j $

For the middle summation we can use Gauss' rule.

$ \sum\limits_{i=1}^{n/2} \frac{1}{2} * (n - i) * (n - i + i) = \sum\limits_{i=1}^{n/2} \frac{1}{2} * (n - i) * (n) $

We can rule out the constants.

$ n*\frac{1}{2}*\sum\limits_{i=1}^{n/2} (n - i) $

We can now use the linearity of summations to get simpler ones.

$ n*\frac{1}{2}*[\sum\limits_{i=1}^{n/2} n - \sum\limits_{i=1}^{n/2} i] $

In the first summation n is just a constant so we can bring it outside and the for the second summation we use, once again, Gauss' rule. The first summation turns into $ \frac{n}{2} $.

The rest is just numerical calculations. The problem is that I cannot get to the answer on the main page: $ \frac{n^3}{8} $. Could somebody help me catch the error? Thank you in advance.


--Mardurhack (talk) 19:45, 12 December 2014 (EST)