In the expansion of (9x-13x)^18,x0, if the term independent of x is (221)k, then k is equ…

Question

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

Accepted Answer

For the expansion of

${(9x-\frac{1}{3\sqrt{x}})}^{18},$

we use the general term in the binomial expansion:

${T}_{r+1}=(\frac{18}{r})(9x{)}^{18-r}{(-\frac{1}{3\sqrt{x}})}^{r}$

Now simplify the power of $x$ in this term.

Since

$(9x{)}^{18-r}={9}^{18-r}{x}^{18-r}$

and

${(\frac{1}{3\sqrt{x}})}^{r}=\frac{1}{{3}^{r}{x}^{r/2}},$

so the power of $x$ in the general term is

${x}^{18-r-\frac{r}{2}}={x}^{18-\frac{3r}{2}}.$

For the term independent of $x$, the power of $x$ must be zero:

$18-\frac{3r}{2}=0$

$\frac{3r}{2}=18$

$r=12$

So the required term is the $(12+1)$th term, i.e. ${T}_{13}$.

Now compute it:

${T}_{13}=(\frac{18}{12})(9x{)}^{6}{(-\frac{1}{3\sqrt{x}})}^{12}$

Since the power of $x$ becomes zero, only the numerical coefficient remains:

${T}_{13}=(\frac{18}{12})\cdot {9}^{6}\cdot \frac{1}{{3}^{12}}$

Also,

${9}^{6}=({3}^{2}{)}^{6}={3}^{12}$

So,

${T}_{13}=(\frac{18}{12})$

and

$(\frac{18}{12})=(\frac{18}{6})=\frac{18!}{6!\, 12!}=18564$

Given that the term independent of $x$ is $(221)k$, we have

$221k=18564$

Thus,

$k=\frac{18564}{221}=84$

So the correct answer is:

$84$

Hence, Option A is correct.

Solutions · 1