Close form of $1+\sum_{j=1}^n j! j$
Moderator: elim
Close form of $1+\sum_{j=1}^n j! j$
by elim » Tue Nov 15, 2011 8:21 pm
Find the close form of $\displaystyle{1+\sum_{j=1}^n j! j }$
- elim
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Re: Close form of $1+\sum_{j=1}^n j! j$
by elim » Tue Nov 15, 2011 8:25 pm
$\displaystyle{1+\sum_{j=1}^n j! j = 1+\sum_{j=1}^n \left((j+1)! -j! \right ) = \sum_{j=1}^{n+1} j! - \sum_{j=1}^n j! = (n+1)! }$
- elim
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- Joined: Mon Feb 08, 2010 8:03 pm
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