Triple Reciprocal Factorial Sum

Let the required sum be $S$ .

$\begin{aligned}S&=\sum_{r=1}^{2010} \dfrac{r+2}{r!+(r+1)!+(r+2)!}\\&=\sum_{r=1}^{2010}\dfrac{r+2}{r!(1+r+1+(r+1)(r+2))}\\&=\sum_{r=1}^{2010}\dfrac{r+2}{r!(r^2+4r+4)}\\&=\sum_{r=1}^{2010}\dfrac{1}{r!(r+2)}\\&=\sum_{r=1}^{2010}\dfrac{r+1}{(r+2)!}\\&=\sum_{r=1}^{2010}\dfrac{r+2-1}{(r+2)!}\\&=\sum_{r=1}^{2010} \dfrac{1}{(r+1)!}-\dfrac{1}{(r+2)!}\\&=\boxed{\dfrac{1}{2!}-\dfrac{1}{2012!}}\end{aligned}$

Therefore, our answer is $\boxed{2014}$ .

Every term in this sequence can be written as $\frac {1}{(x-1)!} - \frac {1}{x!}$ where $x$ denotes the numerator (number at the top) of the fraction. When writing out each term as the difference of two other terms you can see that all but two terms will cancel each other out. The two which won't cancel out are $\frac {1}{2!}$ and $\frac {1}{2012!}$ . Hence the answer is $2+2012=2014$ .

here is my solution :)