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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Show that the middle term in the expansion of $(x-\large\frac{1}{x})^{2n}$ is $\large\frac{1\times 3\times 5\times..... (2n-1)}{n!}$$\times (-2)^n$

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  • $T_{r+1}=nC_rx^{n-r}b^r$
Clearly the number of terms in the expansion of $(x-\large\frac{1}{x})^{2n}$ is $2n+1$
So,middle term =$(n+1) ^{th}$ term=$t_{n+1}$
$\Rightarrow t_{n+1}=(-1)^n 2nC_n.x^{2n-n} .(\large\frac{1}{x})^n$
$\Rightarrow (-1)^n 2nC_n.x^n .\large\frac{1}{x^n}$
$\Rightarrow (-1)^{n}\large\frac{(2n)!}{(n!)\times (n!)}$
$\Rightarrow (-1) ^n \large\frac{}{(n!)\times (n!)}$
$\Rightarrow (-1) ^n \large\frac{[1.3.5......(2n-3)(2n-1)][2.4.6.....(2n-2)2(n)]}{(n!)(n!)}$
$\Rightarrow (-1)^n [1.3.5.....(2n-1)\times 2^n \times [1.2.3(n-1).n]$
$\Rightarrow (-1)^n \large\frac{1.3.5.....(2n-1)}{n!}$$\times 2n$
$\Rightarrow \large\frac{1.3.5.....(2n-1)}{n!}$$\times [-2]^n$
Hence proved.
answered Jun 24, 2014 by sreemathi.v

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