Q57 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is

  • A
    $8$
  • B
    $9$
  • Correct answer
    C
    $\frac{80}{9}$
  • D
    $\frac{29}{3}$

Solution

General term of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is $T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r$ $=(2)^{5-r}\left(\frac{-1}{3}\right)^r{ }^5 C_r x^{15-3 r-2 r}$ $=(2)^{5-r}\left(\frac{-1}{3}\right)^r{ }^5 C_r x^{15-5 r}$ For coefficient of $x^5$ put $r=2$ $\therefore$ Coefficient of $x^5=2^3\left(\frac{-1}{3}\right)^2 C_2=\frac{8}{9} \frac{5!}{2!3!}=\frac{80}{9}$

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

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