Proof of Identity, using binomial theorem

• September 16th 2012, 06:32 AM
renatofernandes28
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
• September 16th 2012, 07:43 AM
GJA
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!
• September 16th 2012, 09:54 AM
renatofernandes28
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?
• September 16th 2012, 10:10 AM
Deveno
Re: Proof of Identity, using binomial theorem
Quote:

Originally Posted by renatofernandes28
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?

i would use $(1+x)^2(1+x)^{n-2}$ instead. do you see why?

$\binom{n}{k}$ is the coefficient of $x^k$ in $(1+x)^n$, where do the $x^k$ terms come from in $(1+x)^2(1+x)^{n-2}$ (hint: there are three such terms)?
• September 16th 2012, 10:35 AM
renatofernandes28
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 ?