Question

Write down and simplify $6^{th}$ term in ${(\frac{2x}{3}+\frac{3y}{2})}^9$.

Answer 1

$\begin{array}{c}{\left[\frac{2x}{3}+\frac{3y}{2}\right]}^9\\ 6^{th}termbybinomialexpansion\\ \because T_{r\ell i}{=}^n{C}_r{\left(x\right)}^{n-r}-y^r\left[\because n_{er}=n_{cr-1}\right]\\ for{6}^{th}term\\ \Rightarrow T_{\left(5+1\right)}{=}^9{C}_5.{\left[\frac{2x}{3}\right]}^{\left(9-5\right)}{\left(\frac{3y}{2}\right)}^5\\ \Rightarrow \frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}{\left(\frac{2x}{3}\right)}^4{\left(\frac{3y}{2}\right)}^5=126\times \frac{2}{34}\times \frac{3}{2}{x}^4{y}^5\\ \Rightarrow 63\times 3=189x^4{y}^{4}Ans.\end{array}$