TADM2E 2.36 - Revision history

Matt: Undo revision 792 by EthanGamer (talk)

2020-07-23T13:37:17Z

Undo revision 792 by EthanGamer (talk)

2020-07-14T01:58:36Z

Blanked the page

2018-12-12T07:29:26Z

The previous answer was way off, this must be the correct one.

2016-10-14T18:24:57Z

Make it prettier

2014-12-13T00:35:30Z

Undo revision 250 by Mardurhack (talk)

2014-12-13T00:32:10Z

2014-09-11T18:24:10Z

Recovering wiki

New page

(n^3) / 8

← Older revision		Revision as of 00:35, 13 December 2014
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−	~~<math>\sum\limits_{i=1}^{n/2} \sum\limits_{j=i}^{n-i} \sum\limits_{k=1}^{j} 1</math>~~	+	(n^3) / 8
−
−	~~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>~~

← Older revision		Revision as of 00:32, 13 December 2014
Line 1:		Line 1:
−	(n^3) / 8	+	<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>