If ( 1 + x ) ^n = C 0 + C 1 x + C 2 x 2 + ⋯ + C n x n then show that (C0+C1+C2+...+Cn)^2=2nC0+2nC1+2nC2+...+2nC2n?

1 Answer
Nov 28, 2017

See below.

Explanation:

Assuming that the question reads

If #( 1 + x ) ^n = C_0 + C_1 x + C_2 x^2 + ⋯ + C_n x^n# then show that #(C_0+C_1+C_2+cdots+C_n)^2=2^nC_0+2^nC_1+2^nC_2+cdots+2^nC_n#?

This is trivially answered knowing that with #x = 1#

#2^(2n) = 2^n xx 2^n#