Quickie #6

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December 30th 2006, 02:02 PM
Soroban

Quickie #6

Find the sum of the series:

. . $1(1!) + 2(2!) + 3(3!) + \cdots + n(n!)$
December 30th 2006, 02:43 PM
ThePerfectHacker

Quote:

Originally Posted by Soroban

Find the sum of the series:

. . $1(1!) + 2(2!) + 3(3!) + \cdots + n(n!)$

Quick already solved it.
(n+1)!-1

It is just mathematical iduction
December 30th 2006, 03:41 PM
TD!

I don't know where it was solved, but even without (explicit) induction:

$<br /> k\left( {k!} \right) = \left( {k + 1} \right)k! - k! = \left( {k + 1} \right)! - k!<br />$

So summing for k = 1 till n gives (n+1)!-1 since all intermediate terms cancel out.

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