Difference between revisions of "TADM2E 2.36"
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| − | (n^3 | + | <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 distance in the progression in 1. | ||
| + | |||
| + | <math>\sum\limits_{i=1}^{n/2} \sum\limits_{j=i}^{n-i} j</math> | ||
| + | |||
| + | For the middle summation we can use Gauss' rule. | ||
| + | |||
| + | <math>\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)</math> | ||
| + | |||
| + | We can rule out the constants. | ||
| + | |||
| + | <math>n*\frac{1}{2}*\sum\limits_{i=1}^{n/2} (n - i)</math> | ||
| + | |||
| + | We can now use the linearity of summations to get simpler ones. | ||
| + | |||
| + | <math>n*\frac{1}{2}*[\sum\limits_{i=1}^{n/2} n - \sum\limits_{i=1}^{n/2} i]</math> | ||
| + | |||
| + | 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 <math>\frac{n}{2}</math>. | ||
| + | |||
| + | <math>n^{2}*\frac{1}{2}*[\frac{n}{2} - (\frac{n}{2}*(\frac{n}{2} + 1))*\frac{1}{2}] = \frac{n^3}{16}</math> | ||
Revision as of 00:32, 13 December 2014
$ \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} $
