Proof of Identity, using binomial theorem

I need to prove the following using the binomial theorem

(n,k) = (n-2,k) + 2(n-2,k-1) + (n-2,k-2)

n

The binomial theorem states (1+x)^n = Sum (n,k)x^k

k=0

I don't know where to start, any hints would be helpful! :) ... Also (n,k) = n Choose k incase my text isn't clear

Re: Proof of Identity, using binomial theorem

Hey, renato.

Have you ever seen Pascal's identity? It's a corollary to the Binomial Theorem. It says

C(n,k)=C(n-1,k-1)+C(n-1,k).

This is restating in terms of combinations that the elements of Pascal's triangle are determined by the two elements directly above it. Anyways, if we use this identity on C(n,k), then on C(n-1,k-1) and C(n-1,k) we should get what we want.

Does this get things going in the right direction? Let me know if anything is confusing. Good luck!

Re: Proof of Identity, using binomial theorem

Hmmm i don't think i can use Pascals identity, I'm considering expanding both sides of (1+x)^n = (1+x)(1+x)^(n-1), will that get me somewere?

Re: Proof of Identity, using binomial theorem

Re: Proof of Identity, using binomial theorem

Should i try to expand (1+x)^2(1+x)^(n-2) ? ... I'm trying to get (1+x)^(n-2) + 2(1+x)^n-2 + (1+x)^n-2 ? .... And one more question how does the k term affect the (1+x)^n term? for example how doesnt C(n-2,k-2) look like as a binomal expansion ?