MHB What is the Sum of Factorials for a Specific Equation?

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The discussion focuses on evaluating the sum of factorials for the equation $$\frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}$$. It simplifies the expression using the identity $$\dfrac{n^2-2}{n!}=\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!}$$. The final evaluation leads to the result of $$3-\dfrac {1}{2011!}-\dfrac{2}{2012!}$$. This approach effectively breaks down the sum into manageable parts, highlighting the relationship between factorials and the series. The conclusion provides a clear answer to the original question posed in the thread.
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Evaluate $$\frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}$$
 
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anemone said:
Evaluate $$\frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}$$
$\dfrac{n^2-2}{n!}=\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!}$

$\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!})$

$=3-\dfrac {1}{2011!}-\dfrac{2}{2012!}$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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