The coefficient of frac{1}{x} in the expansio | vedclass

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(D) Given expression is ${\left( {1 + x} \right)^n}{\left( {1 + \frac{1}{x}} \right)^n} = {\left( {1 + x} \right)^n}{\left( {\frac{{x + 1}}{x}} \right

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(D) Given expression is ${\left( {1 + x} \right)^n}{\left( {1 + \frac{1}{x}} \right)^n} = {\left( {1 + x} \right)^n}{\left( {\frac{{x + 1}}{x}} \right)^n} = \frac{{{{\left( {1 + x} \right)}^{2n}}}}{{{x^n}}}$.The general term in the expansion of ${\left( {1 + x} \right)^{2n}}$ is $T_{r+1} = {}^{2n}C_r x^r$.Therefore,the general term in the expansion of $\frac{{{{\left( {1 + x} \right)}^{2n}}}}{{{x^n}}}$ is $\frac{{{}^{2n}C_r x^r}}{{{x^n}}} = {}^{2n}C_r x^{r-n}$.To find the coefficient of $\frac{1}{x}$,we set the exponent $r - n = -1$,which gives $r = n - 1$.The coefficient is ${}^{2n}C_{n-1} = \frac{(2n)!}{(n-1)!(2n - (n-1))!} = \frac{(2n)!}{(n-1)!(n+1)!}$.

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

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