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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? |
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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? |
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Why not just using the basic binomial coeficient identity twice? $$ \binom{n + 1}{k + 1} = \binom{n}{k + 1} + \binom{n}{k} $$ |
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