If the coefficients of the (2r+6)^{text{th}} | vedclass

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(C) The general term in the expansion of $(1+x)^{n}$ is given by $T_{k+1} = {}^{n}C_{k} x^{k}$.For the expansion of $(1+x)^{21}$,the coefficient of th

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$6$

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(C) The general term in the expansion of $(1+x)^{n}$ is given by $T_{k+1} = {}^{n}C_{k} x^{k}$.For the expansion of $(1+x)^{21}$,the coefficient of the $(k+1)^{	ext{th}}$ term is ${}^{21}C_{k}$.The coefficient of the $(2r+6)^{	ext{th}}$ term is ${}^{21}C_{(2r+6)-1} = {}^{21}C_{2r+5}$.The coefficient of the $(r-1)^{	ext{th}}$ term is ${}^{21}C_{(r-1)-1} = {}^{21}C_{r-2}$.Given that these coefficients are equal,we have ${}^{21}C_{2r+5} = {}^{21}C_{r-2}$.Using the property ${}^{n}C_{a} = {}^{n}C_{b}$,which implies either $a = b$ or $a + b = n$:Case $1$: $2r+5 = r-2 \Rightarrow r = -7$ (Not possible as $r$ must be a positive integer).Case $2$: $(2r+5) + (r-2) = 21$ $\Rightarrow 3r + 3 = 21$ $\Rightarrow 3r = 18$ $\Rightarrow r = 6$.

If the coefficients of the $(2r+6)^{\text{th}}$ and $(r-1)^{\text{th}}$ terms in the expansion of $(1+x)^{21}$ are equal,then the value of $r$ is:

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