In the expansion of (1 + x)^{50},the sum of t | vedclass

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(B) We have $(1 + x)^{50} = \sum_{r=0}^{50} {}^{50}C_r x^r = {}^{50}C_0 + {}^{50}C_1 x + {}^{50}C_2 x^2 + {}^{50}C_3 x^3 + \dots + {}^{50}C_{50} x^{50

Vedclass · Accepted Answer

$2^{49}$

Vedclass · Answer

(B) We have $(1 + x)^{50} = \sum_{r=0}^{50} {}^{50}C_r x^r = {}^{50}C_0 + {}^{50}C_1 x + {}^{50}C_2 x^2 + {}^{50}C_3 x^3 + \dots + {}^{50}C_{50} x^{50}$.Let $S_o$ be the sum of coefficients of odd powers of $x$ and $S_e$ be the sum of coefficients of even powers of $x$.$S_o = {}^{50}C_1 + {}^{50}C_3 + \dots + {}^{50}C_{49}$ and $S_e = {}^{50}C_0 + {}^{50}C_2 + \dots + {}^{50}C_{50}$.We know that $(1 + 1)^{50} = S_e + S_o = 2^{50}$ and $(1 - 1)^{50} = S_e - S_o = 0$.Subtracting the two equations: $(S_e + S_o) - (S_e - S_o) = 2^{50} - 0$.$2S_o = 2^{50}$.$S_o = \frac{2^{50}}{2} = 2^{49}$.

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

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