The middle term in the expansion of {left( {1 | vedclass

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Question

(D) The given expression is ${\left( {1 - \frac{1}{x}} \right)^n}{\left( {1 - x} \right)^n}$.This can be rewritten as ${\left( \frac{x-1}{x} \right)^n

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$ {}^{2n}{C_n}$

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(D) The given expression is ${\left( {1 - \frac{1}{x}} ight)^n}{\left( {1 - x} ight)^n}$.This can be rewritten as ${\left( \frac{x-1}{x} ight)^n} {(1-x)^n} = {\left( \frac{-(1-x)}{x} ight)^n} {(1-x)^n} = {(-1)^n} {x^{-n}} {(1-x)^{2n}}$.The expansion of ${(1-x)^{2n}}$ has $2n+1$ terms.The middle term is the ${\left( \frac{2n+1+1}{2} ight)}^{	ext{th}} = {(n+1)}^{	ext{th}}$ term.The general term $T_{r+1}$ in the expansion of ${(1-x)^{2n}}$ is given by ${}^{2n}C_r {(-x)^r} = {}^{2n}C_r {(-1)^r} {x^r}$.For the $(n+1)^{	ext{th}}$ term,we set $r=n$,which gives ${}^{2n}C_n {(-1)^n} {x^n}$.Substituting this back into the expression: ${(-1)^n} {x^{-n}} \cdot {}^{2n}C_n {(-1)^n} {x^n} = {(-1)^{2n}} {}^{2n}C_n = {}^{2n}C_n$ (since $2n$ is even,${(-1)^{2n}} = 1$).

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

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