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Q. Assertion : The term independent of $x$ in the expansion of $\left(x+\frac{1}{x}+2\right)^{m}$ is $\frac{(4 m) !}{(2 m !)^{2}}$
Reason : The coefficient of $x ^{6}$ in the expansion of $(1+ x )^{ n }$ is ${ }^{ n } C _{6}$

Binomial Theorem

Solution:

$\left(x+\frac{1}{x}+2\right)^{m}=\left(\frac{x^{2}+2 x+1}{x}\right)^{m}=\frac{(1+x)^{2 m}}{x^{m}}$
Term independent of $x$ is coefficient of $x^{m}$ in the
expansion of $(1+ x )^{2 m }={ }^{2 m } C _{ m }=\frac{(2 m ) !}{( m !)^{2}}$
Coefficient of $x^{6}$ in the expansion of $(1+x)^{n}$ is ${ }^{n} C_{6}$