In the expansion of (x+a)n, if the sum of odd terms P and sum of even be Q, prove that (i)P2-Q2=(x2-a2)n. (ii) 4PQ=(x+a)2n-(x-a)2n

P^2-Q^2 can be written as (P+Q)(P-Q). P+Q is equal to (x+a)^n and P-Q is equal to nC0×x^n-nC1×x^(n-1)+nC2×x^(n-2)-… or (x-a)^n.

So P^2-Q^2=(x+a)^n×(x-a)^n=(x^2-a^2)^n

(I) part is proved

similarily if we subtract (p –q)^2 from (p+q)^2

we get 4pq which will directly come to (x+a)^2n – (x –a)^2n

(II) part is proved