Question

The middle term in the expansion of ${(\frac{2x^2}{3}+\frac{3}{2x^2})}^{10}$ is

• A. $251$
• B. $252$
• C. $250$
• D. None

Answer 1

The correct option is B $252$

$(\frac{2x^2}{3}+\frac{3}{2x^2}{)}^{10}$

$\Rightarrow (a+b{)}^n{=}^n{C}_0{a}^n{+}^{n}4a^{n}b+--{+}^n{C}_{\pi }{a}^{n-\pi }{b}^{\pi }+{-}^n{C}_n{b}^n$

$\therefore$ middle term is $6^{th}$ here, so :

$=7_6{\Rightarrow }^{10}{C}_5\left(\frac{2x^2}{3}{)}^{10-5}\right(\frac{2}{2x^2}{)}^5$

$\Rightarrow \frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 5\times 2\times 1}\left(\frac{2x^2}{3^8}{)}^5\right(\frac{2}{2x^2}{)}^5$

$\Rightarrow 4\times 63\times 2\left(1\right)$

$\Rightarrow 252\times 1$

$\Rightarrow 252$

(Option B)