- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,937

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

Last edited:

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,937

Last edited:

- Jan 25, 2013

- 1,225

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

$\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!})$

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