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Q. The coefficient of $1 / x$ in the expansion of $(1+ x )^{ n }(1+$ $1 / x )$ is

Binomial Theorem

Solution:

Coefficient of $x ^{-1}$ in $(1+ x )^{ n }\left(1+\frac{1}{ x }\right)^{ n }$
$=$ Coefficient of $x ^{-1}$ in $\frac{(1+ x )^{2 n }}{ x ^{ n }}$
$=$ Coefficient of $x ^{ n -1}$ in $(1+ x )^{2 n }$
$={ }^{2 n } C _{ n -1}=\frac{(2 n ) !}{( n -1) !( n +1) !}$