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> How to find the summation of $\Sigma\frac{(n+1)(n+2)}{2!}$ ?
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**MY WORK:**

I know that the expression in the summation is the general term of the binomial expansion $(1-x)^{-3}$ . I have a solution where I consider :
$$(1-x)^{-3}=a\_0+a\_1x+a\_2x^2 +...+a\_nx^n +...$$
Then I multiply it with the following :
$$(1-x)^{-1}=1+x+x^2+x^3+ ... + x^n+ ...$$
Equating co efficients of $x^n$ , I get :
$$a\_0+a\_1+a\_2+...+a\_n=\Sigma\frac{(n+1)(n+2)}{2!}=\frac{(n+1)(n+2)(n+3)}{3!}$$ ... 

**CAN I DO IT IN ANY OTHER WAY OTHER THAN BINOMIAL EXPANSION?**