Close form of $1+\sum_{j=1}^n j! j$

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Close form of $1+\sum_{j=1}^n j! j$

Postby 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 $1+\sum_{j=1}^n j! j$

Postby elim » Tue Nov 15, 2011 8:25 pm

$\displaystyle{1+\sum_{j=1}^n j! j = 1+\sum_{j=1}^n \left((j+1)! -j! \right ) = \sum_{j=1}^{n+1} j! - \sum_{j=1}^n j! = (n+1)! }$
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