Question

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

• A. \(42\)
• B. \(64\)
• C. \(72\)
• D. \(84\)

Hint:

For the constant term in binomial expansion, set the power of the variable equal to zero.

Answer 1

The Correct Option is D

Concept: The general term in the binomial expansion \[ (a+b)^n \] is \[ T_{r+1}={}^{n}C_r a^{\,n-r}b^{\,r} \]
Step 1: Identify terms \[ a=9x, \qquad b=-\frac{1}{3\sqrt{x}}, \qquad n=18 \] General term: \[ T_{r+1}={}^{18}C_r(9x)^{18-r}\left(-\frac{1}{3\sqrt{x}}\right)^r \] \[ T_{r+1}={}^{18}C_r\,9^{18-r}\left(-\frac13\right)^r x^{18-r-\frac r2} \]
Step 2: Find term independent of \(x\) Power of \(x\) must be zero: \[ 18-r-\frac r2=0 \] \[ 18=\frac{3r}{2} \] \[ r=12 \]
Step 3: Find coefficient \[ T_{13}={}^{18}C_{12}\,9^{6}\left(\frac13\right)^{12} \] \[ 9^6=(3^2)^6=3^{12} \] Thus \[ 9^6\left(\frac13\right)^{12}=1 \] Hence coefficient \[ {}^{18}C_{12}={}^{18}C_6=18564 \]
Step 4: Find \(k\) Given \[ 221k=18564 \] \[ k=\frac{18564}{221} \] \[ k=84 \]