If (x+1)^n = sum_(k=0)^n c_k x^k then show sum_(k=0)^n 3^k c_k = 2^(2n) ?

1 Answer
Nov 23, 2017

See the proof below

Explanation:

We have

(x+1)^n=sum_(k=0)c_kx^k

Let x=3

Then,

(3+1)^n=sum_(k=0)c_k3^k

4^n=sum_(k=0)c_k3^k

sum_(k=0)3^kc_k=(2^2)^n=2^(2n)

QED