In expansion of {(1 + x)^{50}}, sum of coefficient of odd powers of x is

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7.Binomial Theorem

easy

In the expansion of ${(1 + x)^{50}},$ the sum of the coefficient of odd powers of $x$ is

A

$0$

B

${2^{49}}$

C

${2^{50}}$

D

${2^{51}}$

Solution

(b) We have, ${(1 + x)^{50}} = \sum\limits_{r = 0}^{50} {{}^{50}{C_r}{x^r}} $.

Therefore, sum of coefficients of odd power of

$x = {}^{50}{C_1} + {}^{50}{C_3} + … + {}^{50}{C_{49}}$

= $\frac{1}{2}[{}^{50}{C_0} + {}^{50}{C_1} + … + {}^{50}{C_{50}}]\,\, = \,\,\frac{1}{2}[{2^{50}}] = {2^{49}}$.

Standard 11

Mathematics

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