Coefficient of x^5 in expansion of left(2 x^3-frac{1}{3 x^2}right)^5 is

English

Gujarati

Hindi

7.Binomial Theorem

medium

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

A

$8$

B

$9$

C

$\frac{80}{9}$

D

$\frac{26}{3}$

(JEE MAIN-2023)

Solution

$\left(2 x^3-\frac{1}{3 x^2}\right)^5$

$T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r}$

$\therefore 15-5 r =5$

$\therefore r =2$

$T_3=10\left(\frac{8}{9}\right) x^5$

So, coefficient is $\frac{80}{9}$

Standard 11

Mathematics

Share

Similar Questions

In the expansion of ${\left( {3x – \frac{1}{{{x^2}}}} \right)^{10}}$ then $5^{th}$ term from the end is :-

normal

In the expansion of ${\left( {x – \frac{3}{{{x^2}}}} \right)^9},$ the term independent of $x$ is

easy

Let the coefficients of third, fourth and fifth terms in the expansion of $\left(x+\frac{a}{x^{2}}\right)^{n}, x \neq 0,$ be in the ratio $12: 8: 3 .$ Then the term independent of $x$ in the expansion, is equal to …… .

medium

(JEE MAIN-2021)

If the second, third and fourth terms in the expansion of $(x+y)^{\mathrm{n}}$ are $135$,$30$ and $\frac{10}{3}$, respectively, then $6\left(n^3+x^2+y\right)$ is equal to ………….

hard

(JEE MAIN-2024)

The coefficient of $x^{2012}$ in the expansion of $(1-x)^{2008}\left(1+x+x^2\right)^{2007}$ is equal to

hard

(JEE MAIN-2024)