Statement-1: The coefficient of the term of independent of x in the expansion of x+9x+6n is 3n(2n)!n!n!Statement-2: The coefficient of xr in the expansion of (1+x)n is nr

# Statement-1: The coefficient of the term of independent of x in the expansion of ${\left(x+\frac{9}{x}+6\right)}^{n}$ is $\frac{{3}^{n}\left(2n\right)!}{n!n!}$Statement-2: The coefficient of xr in the expansion of is $\left(\begin{array}{l}n\\ r\end{array}\right)$

1. A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

2. B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

3. C

STATEMENT-1 is True, STATEMENT-2 is False

4. D

STATEMENT-1 is False, STATEMENT-2 is True

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### Solution:

${\left(x+\frac{9}{x}+6\right)}^{n}={\left(\frac{{x}^{2}+6x+9}{x}\right)}^{n}=\frac{\left(3+x{\right)}^{2n}}{{x}^{n}}$

$=\frac{{3}^{2n}}{{x}^{n}}{\left(1+\frac{x}{3}\right)}^{2n}$

$\therefore$Coefficient of the term independent of $x{\left(x+\frac{9}{x}+6\right)}^{n}$

$={3}^{2n}\left(\begin{array}{c}2n\\ n\end{array}\right){\left(\frac{1}{3}\right)}^{n}=\frac{{3}^{n}\left(2n\right)!}{n!n!}$                    [Using statement -2]

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