Question

The term independent of $x$ in the expansion of $\left(x+\frac{1}{x^{2}}\right)^{6}$ is

• A. 20
• B. 15
• C. 6
• D. 1
• E. 0

Answer 1

Answer: (B)

The general term of $\left(x+\frac{1}{x^{2}}\right)^{6}$ is
$T_{r+1}={ }^{6} C_{r} x^{r}\left(\frac{1}{x^{2}}\right)^{6-r}$
$={ }^{6} C_{r} x^{r-12+2 r}$
For independent of $x$,
$r-12+2 r=0 $
$\Rightarrow 3 r=12 $
$\Rightarrow r=4$
$\therefore $ Required term $={ }^{6} C_{4}=\frac{6 \times 5}{2 \times 1}=15 $