If the coefficients of the (2r+1)^{text{th}} | vedclass

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(B) The general term in the expansion of $(1+x)^n$ is given by $T_{k+1} = {^nC_k} x^k$. For the expansion of $(1+x)^{42}$,the coefficients are: Coeffi

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$14$

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(B) The general term in the expansion of $(1+x)^n$ is given by $T_{k+1} = {^nC_k} x^k$. For the expansion of $(1+x)^{42}$,the coefficients are: Coefficient of $(2r+1)^{	ext{th}}$ term is ${^{42}C_{2r}}$. Coefficient of $(r+1)^{	ext{th}}$ term is ${^{42}C_r}$. Given that these coefficients are equal: ${^{42}C_{2r}} = {^{42}C_r}$. Using the property ${^nC_x} = {^nC_y}$,we have two cases: Case $1$: $x = y$ $\Rightarrow 2r = r$ $\Rightarrow r = 0$. Case $2$: $x + y = n$ $\Rightarrow 2r + r = 42$ $\Rightarrow 3r = 42$ $\Rightarrow r = 14$. Since $r$ must be a positive integer in this context,$r = 14$ is the valid solution.

If the coefficients of the $(2r+1)^{\text{th}}$ term and the $(r+1)^{\text{th}}$ term in the expansion of $(1+x)^{42}$ are equal,then $r$ can be

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