<s>[INST] Derive the generating function: $\frac1{(1-x)^3}= \sum_{n=0}^\infty \binom{n+2}{2}x^n$ 
$$\frac1{(1-x)^3} = \sum_{n=0}^\infty \binom{n+2}2x^n$$

I am not sure how to begin.
$$\begin{aligned}\frac1{1-x}&= \sum_{n=0}^{\infty} x^n\\\frac1{(1-x)^2}&=\sum_{n=0}^{\infty} nx^{n-1}\end{aligned}$$
Then you would get:
$$\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} nx^{2n - 1}$$
Which is nowhere near the same...?
 [/INST] Take the derivative of both sides
$$\frac{d}{dx} \left (\frac{1}{(1-x)^2} \right) = \frac{d}{dx} \left ( \sum_{n=0}^{\infty} n x^{n-1} \right)$$
$$-(-2)\frac{1}{(1-x)^3}  = \sum_{n=0}^{\infty}  n (n-1) x^{n-2}$$
$$\frac{1}{(1-x)^3}  = \sum_{n=0}^{\infty}  \frac{n (n-1)}{2} x^{n-2} = \sum_{n=0}^{\infty}  \frac{(n +2) (n+1)}{2} x^{n}.$$
Look at the definition of the binomial coefficient and you'll see that this is the same as the desired formula.
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