Question

The term independent of x in the expansion of ${\left(2x+\frac{1}{3x^2}\right)}^9$ is

• A. $2^{nd}$
• B. $3^{rd}$
• C. $4^{th}$
• D. $5^{th}$

Answer 1

Answer: (C)

Suppose (r + 1)th term is independent of x.
We have
$T_{r+1}{=}^9{C}_r{\left(2x\right)}^{9-r}{\left(\frac{1}{3x^2}\right)}^r$
${=}^9{C}_r{2}^{9-r}\frac{1}{3^r}.x^{9-3r}$
This term is independent of x if 9 - 3r = 0
i.e., r = 3.
Thus, 4th term is independent of x.