Using Binomial Theorem to prove identity I need to prove the following using the binomial theorem
$${n \choose k} = {n-2 \choose k} + 2{n-2 \choose k-1} + {n-2 \choose k-2}$$
The binomial theorem states $$(1+x)^n = \sum_{k=0}^n {n \choose k} x^k$$
I'm trying to use $(1+x)^n = (1+x)^2(1+x)^{n-2}$ but i dont know how to go from there?
 A: You're quite close already - note that by the Binomial Theorem, $$(1+x)^{n-2} = \sum_{k=0}^{n-2} {n-2 \choose k} x^k.$$
Expanding $(1+x)^2 = 1 + 2x + x^2$, we get
$$(1+x)^n = (1+x)^2(1+x)^{n-2} = \sum_{k=0}^{n-2} {n-2 \choose k} \left(x^k +2 x^{k+1} +x^{k+2}\right).$$
Now, the trick is to collect the different powers of $x$, that iş tranforming the formula in a way that only $x^k$ appears on the right-hand side. Can you do this?
A: Why not just using the basic binomial coeficient identity twice?
$$
\binom{n + 1}{k + 1} = \binom{n}{k + 1} + \binom{n}{k}
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{{n \choose k} = {n - 2 \choose k} + 2{n - 2 \choose k - 1}
     +{n - 2 \choose k - 2}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large%
{n - 2 \choose k} + 2{n - 2 \choose k - 1} + {n - 2 \choose k - 2}}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}\bracks{%
{\pars{1 + z}^{n - 2} \over z^{k + 1}}
+2\,{\pars{1 + z}^{n - 2} \over z^{k}}
+{\pars{1 + z}^{n - 2} \over z^{k - 1}}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n - 2} \over z^{k + 1}}
\pars{1 + 2z + z^{2}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z^{k + 1}}
\,{\dd z \over 2\pi\ic} = \color{#66f}{\large{n \choose k}}
\end{align}

