Binomial (1 Viewer)

[currysauce]

I've done this, jst thought i'd let you's have a crack for .... fun

By considering (1-x)^n (1 +1/x)^n

express,

(n 2)(n 0) - (n 3) (n 1) +...+ (-1)^n (n n) (n n-2) where (n n) = nCk

in simplest form

 

[KFunk]

Hmm, my geuss will be 0 if n is odd and ⁿC_(n-2)/2.(-1)^(n+2)/2 if n is even.

 

[LostAuzzie]

(1 - x)^n(1 + 1/x)^n = (1/x - x)^n
LHS = [nC0 - nC1x + nC2x^2 + ...][nCo + nC1(1/x) + nC2(1/x^2) + ...]
RHS = SUM (nCr) (1/x)^(n-r)x^r
= SUM (nCr) x^(-n+r)x^r
= SUM (nCr) x^(2r-n)
Considering coefficients of x^2 on either side:
LHS: (nC2)(nC0) - (nC1)(nC3) + ... + (-1)^n(nCn)(nC(n-2))
RHS: 2r-n = 2
r = (n + 2)/2
Coeff x^2 on RHS: nC((n+2)/2)
Therefore (nC2)(nC0) - (nC3) (nC1) +...+ (-1)^n (nCn) (nC(n-2))