If ${(x + 1)}^{n} = n \sum k = 0 c_{k} x^{k}$ $(x+1)^n = sum_(k=0)^n c_k x^k$ then show $n \sum k = 0 3^{k} c_{k} = 2^{2 n}$ $sum_(k=0)^n 3^k c_k = 2^(2n)$ ?

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Answer 1 · 2017-11-23T11:39:00

See the proof below

We have

${(x + 1)}^{n} = \sum k = 0 c_{k} x^{k}$ $(x+1)^n=sum_(k=0)c_kx^k$

Let $x = 3$ $x=3$

Then,

${(3 + 1)}^{n} = \sum k = 0 c_{k} 3^{k}$ $(3+1)^n=sum_(k=0)c_k3^k$

$4^{n} = \sum k = 0 c_{k} 3^{k}$ $4^n=sum_(k=0)c_k3^k$

$\sum k = 0 3^{k} c_{k} = {(2^{2})}^{n} = 2^{2 n}$ $sum_(k=0)3^kc_k=(2^2)^n=2^(2n)$

$Q E D$ $QED$

If (x+1)^n = sum_(k=0)^n c_k x^k(x+1)n=n∑k=0ckxk(x+1)^n = sum_(k=0)^n c_k x^k then show sum_(k=0)^n 3^k c_k = 2^(2n)n∑k=03kck=22nsum_(k=0)^n 3^k c_k = 2^(2n) ?