Question

Find the term in ${(\sqrt[3]{3\frac{a}{\sqrt{b}}}+\sqrt{\frac{b}{\sqrt[3]{3a}}})}^{21}$ which has the same power of a and b.

Answer 1

$T_{r+1}={}^n{C}_r{\left(\sqrt[3]{3\left(\frac{a}{\sqrt{b}}\right)}\right)}^{n-r}{\left(\sqrt[3]{3\left(\frac{b}{\sqrt{a}}\right)}\right)}^r={}^n{C}_r\left(\frac{a^{\left(\frac{n-r}{3}\right)}}{b^{\left(\frac{n-r}{6}\right)}}\right)\left(\frac{b^{\left(\frac{r}{2}\right)}}{a^{\left(\frac{r}{6}\right)}}\right)={}^n{C}_r{a}^{\left(\right(\frac{n}{3})-(\frac{r}{2}\left)\right)}{b}^{\left(\right(\frac{-n}{6})+(\frac{2r}{3}\left)\right)}nowasperquestion\left(\frac{n}{3}\right)-\left(\frac{r}{2}\right)=\left(\frac{-n}{6}\right)+\left(\frac{2r}{3}\right)\left(\frac{21}{3}\right)-\left(\frac{r}{2}\right)=\left(\frac{-21}{6}\right)+\left(\frac{2r}{3}\right)\left(\frac{r}{2}\right)+\left(\frac{2r}{3}\right)=\left(\frac{21}{3}\right)+\left(\frac{21}{6}\right)\left(\frac{3r+4r}{6}\right)=\left(\frac{42+21}{6}\right)7r=63r=9\therefore term=r+1={10}^{th}term$