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(C) In the expansion of $(1 + x)^{2n + 2}$,the total number of terms is $2n + 3$,which is odd. Thus,the middle term is the $(\frac{2n + 2}{2} + 1)^{th

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$p = q + r$

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(C) In the expansion of $(1 + x)^{2n + 2}$,the total number of terms is $2n + 3$,which is odd. Thus,the middle term is the $(\frac{2n + 2}{2} + 1)^{th} = (n + 2)^{th}$ term.Its coefficient is $p = {}^{2n + 2}C_{n + 1}$.In the expansion of $(1 + x)^{2n + 1}$,the total number of terms is $2n + 2$,which is even. Thus,the middle terms are the $(n + 1)^{th}$ and $(n + 2)^{th}$ terms.Their coefficients are $q = {}^{2n + 1}C_n$ and $r = {}^{2n + 1}C_{n + 1}$.Using the Pascal's identity ${}^{n}C_r + {}^{n}C_{r - 1} = {}^{n + 1}C_r$,we have:$q + r = {}^{2n + 1}C_n + {}^{2n + 1}C_{n + 1} = {}^{2n + 2}C_{n + 1}$.Therefore,$p = q + r$.

If the coefficient of the middle term in the expansion of $(1 + x)^{2n + 2}$ is $p$ and the coefficients of the two middle terms in the expansion of $(1 + x)^{2n + 1}$ are $q$ and $r$,then:

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