Question

The $5$th term in the expansion of ${(2x^2+\frac{3}{x})}^5$ is

• A. $810.x^{-2}$
• B. $810.x^{-4}$
• C. $810$
• D. $810.x^{-5}$

Answer 1

The correct option is A $810.x^{-2}$

we k.n.t $(r+1{)}^{th}$ term in expansion of $(a+b{)}^n$ is ${}^n{c}_r{a}^{n-r}{b}^r$

$\Rightarrow T_{r+1}{=}^n{c}_r{a}^{n-r}{b}^r$

Here $a=2x^2$ and $b=\frac{2}{x}$ and $n=5$

5th term = $T_5{=}^5{c}_4{a}^{5-4}{b}^4=5a{b}^4=5\left(2x^2\right)(\frac{3}{x}{)}^4=(5\ast 2\ast 3^4\left)\right(x^{2-4}=810x^{-2}$

$\therefore$ The $5$th term in expansion of $(2x^2+\frac{3}{x}{)}^4$ is $810x^{-2}$.