The middle term in the expansion of {left( {x | vedclass

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(B) For the expansion of $(x + a)^{2n}$,the total number of terms is $2n + 1$,which is odd. Therefore,the middle term is the $(n+1)$-th term.Using the

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$\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}{n!}$

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(B) For the expansion of $(x + a)^{2n}$,the total number of terms is $2n + 1$,which is odd. Therefore,the middle term is the $(n+1)$-th term.Using the general term formula $T_{r+1} = {}^{2n}C_r (x)^{2n-r} (\frac{1}{2x})^r$,for the middle term,we set $r = n$.$T_{n+1} = {}^{2n}C_n (x)^{2n-n} (\frac{1}{2x})^n = {}^{2n}C_n \cdot x^n \cdot \frac{1}{2^n x^n} = \frac{(2n)!}{n! n! 2^n}$.Expanding $(2n)! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot (2n)$,we separate the odd and even terms:$(2n)! = [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdot \dots \cdot 2n] = [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot 2^n \cdot [1 \cdot 2 \cdot 3 \cdot \dots \cdot n] = [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot 2^n \cdot n!$.Substituting this back into the expression:$T_{n+1} = \frac{[1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot 2^n \cdot n!}{n! \cdot n! \cdot 2^n} = \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{n!}$.

The middle term in the expansion of ${\left( {x + \frac{1}{{2x}}} \right)^{2n}}$ is:

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