In the expansion of left(9x - frac{1}{3sqrt{x | vedclass

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(A) The general term in the expansion of $(a + b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For the expansion $\left(9x - \frac{1}{3\sqrt{x

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$84$

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(A) The general term in the expansion of $(a + b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For the expansion $\left(9x - \frac{1}{3\sqrt{x}}ight)^{18}$, the general term is $T_{r+1} = \binom{18}{r} (9x)^{18-r} \left(-\frac{1}{3}x^{-1/2}ight)^r$. Simplifying the expression, we get $T_{r+1} = \binom{18}{r} 9^{18-r} (-1/3)^r x^{18-r-r/2}$. For the term to be independent of $x$, the exponent of $x$ must be $0$: $18 - \frac{3r}{2} = 0$. Solving for $r$, we get $\frac{3r}{2} = 18$, which implies $r = 12$. Substituting $r = 12$ into the constant part: $\binom{18}{12} 9^{18-12} (-1/3)^{12} = \binom{18}{6} (3^2)^6 (1/3)^{12} = \binom{18}{6} (3^{12}) (1/3^{12}) = \binom{18}{6}$. Calculating the value: $\binom{18}{6} = \frac{18 	imes 17 	imes 16 	imes 15 	imes 14 	imes 13}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = 18564$. Given that the constant term is $(221)k$, we have $221k = 18564$. Thus, $k = \frac{18564}{221} = 84$.

In the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}, x > 0$, if the term independent of $x$ is $(221)k$, then $k$ is equal to:

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