Question

Th coefficient of $x^2$ in the expansion of $\frac{2+x}{{(3-2x)}^2}$ , equals

• A. $\frac{-16}{9}$
• B. $\frac{16}{27}$
• C. $-\frac{1}{4}$
• D. $-\frac{9}{16}$

Answer 1

The correct option is B $\frac{16}{27}$

$(2+x)\frac{1}{9}(1-\frac{2x}{3}{)}^{-2}$
$=\frac{1}{9}(2+x)(1+\frac{4x}{3}+\frac{12x^2}{9}+...\infty )$
Hence coefficient of $x^2$ will be
$\frac{1}{9}[\frac{24}{9}+\frac{4}{3}+\frac{12}{9}]$
$=\frac{1}{9}\left[\frac{48}{9}\right]$
$=\frac{16}{27}$