calculate-k-1-n-kC-n-k-3-k-interms-of-n-

Question Number 64651 by mathmax by abdo last updated on 20/Jul/19

c a l c u l a t e \sum_{k = 1}^{n} {k C}_{n}^{k} 3^{k} i n t e r m s o f n

Commented by mathmax by abdo last updated on 20/Jul/19

l e t p (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k} w e h a v e p (x) = {(x + 1)}^{n} a n d

p^{^{'}} (x) = \sum_{k = 1}^{n} k C_{n}^{k} x^{k - 1} \Rightarrow {x p}^{^{'}} (x) = \sum_{k = 1}^{n} k C_{n}^{k} x^{k}

x = 3 \Rightarrow 3 p^{^{'}} (3) = \sum_{k = 1}^{n} k C_{n}^{k} 3^{k} b u t p^{^{'}} (x) = n {(x + 1)}^{n - 1} \Rightarrow

p^{^{'}} (3) = n 4^{n - 1} \Rightarrow \sum_{k = 1}^{n} {k C}_{n}^{k} 3^{k} = 3 n 4^{n - 1} .

Answered by mr W last updated on 20/Jul/19

{(1 + x)}^{n} = \underset{k = 0}{\sum^{n}} C_{n}^{k} x^{k}

n {(1 + x)}^{n - 1} = \underset{k = 0}{\sum^{n}} {k C}_{n}^{k} x^{k - 1}

n {(1 + x)}^{n - 1} x = \underset{k = 0}{\sum^{n}} {k C}_{n}^{k} x^{k}

w i t h x = 3 :

n {(1 + 3)}^{n - 1} 3 = \underset{k = 0}{\sum^{n}} {k C}_{n}^{k} 3^{k} = \underset{k = 1}{\sum^{n}} {k C}_{n}^{k} 3^{k}

\Rightarrow \underset{k = 1}{\sum^{n}} {k C}_{n}^{k} 3^{k} = 3 n 4^{n - 1}

Commented by mathmax by abdo last updated on 20/Jul/19

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