Question

Find the coefficient of $x^{-2}$ in the expansion of ${(\frac{2x}{3}+\frac{3}{2x})}^6$.

Answer 1

$\begin{array}{c}{\left(\frac{2}{3}+\frac{3}{2x}\right)}^{6}findterms{x}^{-2}\\ \Rightarrow {}^6{C}_0{\left(\frac{2x}{3}\right)}^6{+}^6{C}_1{\left(\frac{2x}{3}\right)}^5{+}^6{C}_2{\left(\frac{2x}{3}\right)}^4{\left(\frac{3}{2x}\right)}^2{+}^6{C}_3{\left(\frac{2x}{3}\right)}^3{\left(\frac{3}{2x}\right)}^3{+}^6{C}_4{\left(\frac{2x}{3}\right)}^2{\left(\frac{3}{2x}\right)}^4{+}^6{C}_5{\left(\frac{2x}{3}\right)}^1{\left(\frac{3}{2x}\right)}^5{+}^6{C}_6{\left(\frac{3}{2x}\right)}^6\\ \Rightarrow \frac{64}{729}{x}^6+\frac{32}{343}{x}^5\cdot \frac{3}{2x}+3c\cdot \frac{16}{81}{x}^4\cdot \frac{9}{4x^2}+120\cdot \frac{8}{27}{x}^3\cdot \frac{27}{8x^3}+360\cdot \frac{4}{9}{x}^2\cdot \frac{81}{16}{x}^4\\ \Rightarrow butwefindtermsof{x}^{-2}=360\cdot \frac{4}{9}\cdot \frac{81}{16}=810\end{array}$