Show that the middle term in the expansion of $(1+x)^{2 n}$ is
$\frac{1.3 .5 \ldots(2 n-1)}{n !} 2 n\, x^{n},$ where $n$ is a positive integer.
Show that the middle term in the expansion of $(1+x)^{2 n}$ is
$\frac{1.3 .5 \ldots(2 n-1)}{n !} 2 n\, x^{n},$ where $n$ is a positive integer.
As $2 n$ is even, the middle term of the expansion $(1+x)^{2 n}$ is $\left(\frac{2 n}{2}+1\right)^{\text {th }}$
i.e., $(n+1)^{\text {th }}$ term which is given by,
${T_{n + 1}} = {\,^{2n}}{C_n}{(1)^{2n - n}}{(x)^n} = {\,^{2n}}{C_n}{x^n} = \frac{{(2n)!}}{{n!n!}}{x^n}$
$=\frac{2 n(2 n-1)(2 n-2) \ldots 4.3 .2 .1}{n ! n !} x^{n}$
$=\frac{1.2 .3 .4 \ldots(2 n-2)(2 n-1)(2 n)}{n ! n !} x^{n}$
$=\frac{[1.3 .5 \ldots(2 n-1)][2.4 .6 \ldots .(2 n)]}{n ! n !} x^{n}$
$=\frac{[1.3 .5 \ldots(2 n-1)] 2^{n}[1.2 .3 \dots n]}{n ! n !} x^{n}$
$=\frac{[1.3 .5 \ldots(2 n-1)] n !}{n ! n !} 2^{n} \cdot x^{n}$
$=\frac{1.3 .5 \ldots(2 n-1)}{n !} 2^{n} x^{n}$
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