So it says both sides but to get the second side try simplify what they gave.

[(1+x)(1-x)]^n = (1 - x^2)^n

Now expand left hand side:

(nC0 + nC1 x + ... + nCn x^n)(nC0 - nC1 x + nC2 x^2 - ... - nCn x^n)

Note that the second brackets has nCn x^n as negative as n is odd.

= [(nC0)^2 - nC0.nC1 x + nC0.nC2 x^2 - .... - nC0.nCn x^n] + [nC1.nC0 x - (nC1)^2 x^2 + ... - nC1.nCn x^(n+1)] + ....

From subbing in x = 1 and expanding you notice every term will cancel out except for the squares of the coefficients, i.e. = (nC0)^2 - (nC1)^2 + ... + (nCn)^2

Expand right hand side and sub in x = 1:

nC0 - nC1 + nC2 - ... - nCn

Since they're the same equation equate both sides:

sigma(k=0 to n) (-1)^k (nCk)^2 = nC0 - nC1 + nC2 - .... - nCn

Via symmetry nC0 = nCn and so on so the right hand side equals 0 if you keep using symmetry and you have your answer

EDIT: Fixed mistake

## Bookmarks