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Algebra

Blazing goIITian

Joined:	29 Mar 2008
Post:	1024

22 Oct 2008 23:10:19 IST

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541

11e

None

Prove that

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Rajat Sen

Blazing goIITian

Joined: 7 Aug 2007

Posts: 533

23 Oct 2008 12:10:30 IST

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let $S = \frac{1.4}{0!} +\frac{2.5}{1!} + .........$

so $S = \sum_{n=0}^{n=\propto}\frac{(n+1)(n+4)}{n!}$

= $\sum\frac{n^{2}}{n!} +\sum\frac{5}{(n-1)!} +\sum\frac{4}{n!}$

= $5e+4e+ \sum\frac{n}{(n-1)!} = 9e + \sum\frac{1}{(n-2)!} +\sum\frac{1}{(n-1)!}$

= $9e+2e = 11e$

please rate! Rolling Eyes

10904him

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Joined: 29 Dec 2006

Posts: 191

23 Oct 2008 12:30:15 IST

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Well done Rjatsen

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