Q67 / 170 · Manipal (MET) 2021 · Mathematics

The term independent of $x$ in the expansion of $\left(2 x^4-\frac{1}{x^2}\right)^{12}$ is

  • A
    6920
  • Correct answer
    B
    7920
  • C
    7900
  • D
    3960

Solution

The general term is $\begin{aligned} T_{r+1} & ={ }^{12} C_r\left(2 x^4\right)^r\left(-\frac{1}{x^2}\right)^{12-r} \\ & ={ }^{12} C_r(2)^r x^{4 r}(-1)^{12-r}(x)^{-2(12-r)}\end{aligned}$ For term independent of $x$, we put $\begin{aligned} & 4 r-24+2 r=0 \Rightarrow 6 r=24 \Rightarrow r=4 \\ & \begin{aligned} T_5 & ={ }^{12} C_4(2)^4(-1)^{12-4} \\ \quad & =\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \times(2)^4=7920\end{aligned}\end{aligned}$

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

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