Question

For $\left|x\right|>\frac{3}{2}$, the value of the third term in the expansion of ${(3+2x)}^{3/5}$ is

• A. $\frac{27}{50}.2^{-\frac{3}{5}}.x^{-\frac{9}{5}}$
• B. $\frac{27}{50}.2^{\frac{3}{5}}.x^{\frac{7}{5}}$
• C. $\frac{27}{50}.2^{-\frac{2}{5}}.x^{-\frac{7}{5}}$
• D. $-\frac{27}{50}.2^{-\frac{2}{5}}.x^{-\frac{7}{5}}$

Answer 1

The correct option is D $-\frac{27}{50}.2^{-\frac{2}{5}}.x^{-\frac{7}{5}}$

Let $y=(2x+3{)}^{3/5}$

on expanding binomially, we get

$y=2^{3/5}{x}^{3/5}+2^{\frac{3}{5}-1}.x^{\frac{3}{5}-1}\times \frac{3}{5}\times 3+(2x{)}^{\frac{3}{5}-2},\frac{3}{5}\times (\frac{3}{5}-1),\frac{1}{2!}\times 3+...$

$\therefore$ The 3rd term = $-2^{-7/5}{x}^{-7/5}\times \frac{27}{25}=-\frac{27}{50}{.2}^{-2/5}{x}^{-7/5}$

$\therefore$ The correct option is (D)