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

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Answer 1 · 2017-11-23T11:39:00

See the proof below

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#