Question

In the expansion of \[ \left(9x-\frac{1}{3\sqrt{x}}\right)^{18}, \] if the coefficient of the term independent of \(x\) is \(221k\), then the value of \(k\) is:

• A. \(82\)
• B. \(83\)
• C. \(84\)
• D. \(86\)

Answer 1

The Correct Option is A

Concept: For the binomial expansion \[ (a+b)^n, \] the general term is \[ T_{r+1}=\binom{n}{r}a^{\,n-r}b^r . \] Step 1: Write the general term. \[ T_{r+1}=\binom{18}{r}(9x)^{18-r}\left(-\frac{1}{3\sqrt{x}}\right)^r \] Step 2: Simplify the powers of \(x\). \[ (9x)^{18-r}=9^{18-r}x^{18-r} \] \[ \left(\frac{1}{3\sqrt{x}}\right)^r =3^{-r}x^{-r/2} \] Thus power of \(x\) becomes \[ x^{18-r-\frac{r}{2}} = x^{18-\frac{3r}{2}} \] Step 3: Find term independent of \(x\). \[ 18-\frac{3r}{2}=0 \] \[ r=12 \] Step 4: Find the coefficient. \[ T_{13}=\binom{18}{12}(9x)^6\left(\frac{1}{3\sqrt{x}}\right)^{12} \] \[ = \binom{18}{12}9^6\frac{1}{3^{12}} \] \[ 9^6=3^{12} \] Hence \[ T_{13}=\binom{18}{12} \] \[ \binom{18}{12}=\binom{18}{6}=18564 \] Step 5: Find \(k\). \[ 18564=221k \] \[ k=84 \]