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- Feb 14, 2012

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Evaluate \(\displaystyle \frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}\)

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- Thread starter
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- Feb 14, 2012

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Last edited:

- Jan 25, 2013

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$\dfrac{n^2-2}{n!}=\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!}$

$\therefore \sum_{2}^{2012}(\dfrac{n^2-2}{n!})=$$ \sum_{2}^{2012}(\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!})$

$=3-\dfrac {1}{2011!}-\dfrac{2}{2012!}$