Difference between revisions of "Talk:TADM2E 2.36"
Mardurhack (Talk  contribs) (Created page with " == Proposed solution (with steps) == <math>\sum\limits_{i=1}^{n/2} \sum\limits_{j=i}^{ni} \sum\limits_{k=1}^{j} 1</math> The innermost summation is just j since the distan...") 
(No difference)

Revision as of 00:45, 13 December 2014
Proposed solution (with steps)
$ \sum\limits_{i=1}^{n/2} \sum\limits_{j=i}^{ni} \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}^{ni} 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)