In the expansion of (1 + x)^n,the sum of the | vedclass

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(D) The binomial expansion is given by $(1 + x)^n = C_0 + C_1x + C_2x^2 + C_3x^3 + C_4x^4 + C_5x^5 + \dots + C_nx^n$.Setting $x = 1$,we get $2^n = C_0

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$2^{n - 1}$

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(D) The binomial expansion is given by $(1 + x)^n = C_0 + C_1x + C_2x^2 + C_3x^3 + C_4x^4 + C_5x^5 + \dots + C_nx^n$.Setting $x = 1$,we get $2^n = C_0 + C_1 + C_2 + C_3 + C_4 + C_5 + \dots + C_n$.Setting $x = -1$,we get $0 = C_0 - C_1 + C_2 - C_3 + C_4 - C_5 + \dots + (-1)^n C_n$.Subtracting the second equation from the first:$2^n - 0 = (C_0 - C_0) + (C_1 - (-C_1)) + (C_2 - C_2) + (C_3 - (-C_3)) + \dots$$2^n = 2(C_1 + C_3 + C_5 + \dots)$.Therefore,the sum of the coefficients of odd powers of $x$ is $C_1 + C_3 + C_5 + \dots = \frac{2^n}{2} = 2^{n - 1}$.

In the expansion of $(1 + x)^n$,the sum of the coefficients of odd powers of $x$ is:

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