# Difference between revisions of "TADM2E 2.36"

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

The innermost summation is just j since the distance in the progression in 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 $\frac{n}{2}$.

$n^{2}*\frac{1}{2}*[\frac{n}{2} - (\frac{n}{2}*(\frac{n}{2} + 1))*\frac{1}{2}] = \frac{n^3}{16}$