C-0-2C-1-3C-2-n-1-C-n-n-2-2-n-1-using-Bionomial-teorm-

Question Number 25969 by soyebshaikh41@gmail.com last updated on 17/Dec/17

C_{0} + {2 C}_{1} + {3 C}_{2} + \dots \dots \dots . + (n + 1) C_{n} = (n + 2) 2^{n - 1}

using Bionomial teorm

Commented by moxhix last updated on 17/Dec/17

s h o w_{n} C_{0} + 2_{n} C_{1} + 3_{n} C_{2} + \dots + (n + 1)_{n} C_{n} = (n + 2) 2^{n - 1}

p u t f (x) = x {(1 + x)}^{n}

f^{'} (x) = {(1 + x)}^{n} + n x {(1 + x)}^{n - 1} = (n + 1 + x) {(1 + x)}^{n - 1}

f^{'} (1) = (n + 2) 2^{n - 1}

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

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

f^{'} (1) = \underset{k = 0}{\sum^{n}}_{n} C_{k} (k + 1)

∴_{n} C_{0} + 2_{n} C_{1} + 3_{n} C_{2} + \dots + (n + 1)_{n} C_{n} = (n + 2) 2^{n - 1}

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