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Question

(B) The expansion of $(1 + x)^m$ has the greatest coefficient at the middle term.For $(1 + x)^{2n + 2}$,the exponent is $m = 2n + 2$,which is an even

Vedclass · Accepted Answer

$\frac{(2n + 2)!}{\{(n + 1)!\}^2}$

Vedclass · Answer

(B) The expansion of $(1 + x)^m$ has the greatest coefficient at the middle term.For $(1 + x)^{2n + 2}$,the exponent is $m = 2n + 2$,which is an even number.The middle term is the $\left(\frac{m}{2} + 1\right)$-th term,which is the $\left(\frac{2n + 2}{2} + 1\right)$-th term = $(n + 2)$-th term.The coefficient of the $(n + 2)$-th term is given by $\binom{2n + 2}{n + 1}$.Thus,the greatest coefficient is $\binom{2n + 2}{n + 1} = \frac{(2n + 2)!}{(n + 1)!(2n + 2 - (n + 1))!} = \frac{(2n + 2)!}{\{(n + 1)!\}^2}$.

The greatest coefficient in the expansion of $(1 + x)^{2n + 2}$ is

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