Q205 / 270 · NDA 2012 (Phase 1) · Mathematics

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

  • A
    9
  • B
    18
  • C
    48
  • Correct answer
    D
    84

Solution

$\begin{aligned} \left(x^2-\frac{1}{x}\right)^9 & \\ t_{r+1} & ={ }^9 C_r\left(x^2\right)^{9-r}\left(\frac{-1}{x}\right)^r \\ & ={ }^9 C_r(x)^{18-2 r}{ }_{(-1)^r x^{-r}} \end{aligned}$ Put $18-3 r=0$ $r=6$ Put $r=6$, in (1), $t_7={ }^9 C_6(1)^6=\frac{9!}{6!3!}=\frac{9 \cdot 8 \cdot 7 \cdot 6!}{6!\cdot 3 \cdot 2}=84 .$

Hence, the correct answer is option: 'D'.

Relevant Questions