Let the coefficients of powers of x in the 2n | vedclass

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(B) The coefficients of the $2nd$, $3rd$, and $4th$ terms in the expansion of $(1+x)^{n}$ are ${}^{n}C_{1}$, ${}^{n}C_{2}$, and ${}^{n}C_{3}$ respecti

Vedclass · Accepted Answer

$64$

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(B) The coefficients of the $2nd$, $3rd$, and $4th$ terms in the expansion of $(1+x)^{n}$ are ${}^{n}C_{1}$, ${}^{n}C_{2}$, and ${}^{n}C_{3}$ respectively.Given that these are in arithmetic progression, we have $2({}^{n}C_{2}) = {}^{n}C_{1} + {}^{n}C_{3}$.Substituting the values, $2 	imes \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6}$.Dividing by $n$ (since $n 
eq 0$), we get $n-1 = 1 + \frac{(n-1)(n-2)}{6}$.$6(n-2) = (n-1)(n-2)$.Since $n 
eq 2$, we divide by $(n-2)$ to get $6 = n-1$, which implies $n = 7$.The sum of the coefficients of odd powers of $x$ in the expansion of $(1+x)^{n}$ is given by $2^{n-1}$.Substituting $n=7$, the sum is $2^{7-1} = 2^{6} = 64$.

Let the coefficients of powers of $x$ in the $2nd$, $3rd$, and $4th$ terms in the expansion of $(1+x)^{n}$, where $n$ is a positive integer, be in arithmetic progression. Then, the sum of the coefficients of odd powers of $x$ in the expansion is

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