Question

The coefficient of the term independent of x in the expansion of ${\left(ax+\frac{b}{x}\right)}^{14}$ is
(a) $14!a^7{b}^7$

(b) $\frac{14!}{7!}{a}^7{b}^7$

(c) $\frac{14!}{{\left(7!\right)}^2}{a}^7{b}^7$

(d) $\frac{14!}{{\left(7!\right)}^3}{a}^7{b}^7$

Answer 1

(c) $\frac{14!}{{\left(7!\right)}^2}{a}^7{b}^7$

$$\begin{gathered} \mathrm{Suppose}(r+1)\mathrm{th}\mathrm{term}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}\mathrm{is}\mathrm{independent}\mathrm{of}x. \\ \mathrm{Then},\mathrm{we}\mathrm{have} \\ {T}_{r+1}={}^{14}C_r(ax{)}^{14-r}{\left(\frac{b}{x}\right)}^r \\ ={}^{14}C_r{a}^{14-r}{b}^r{x}^{14-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have} \\ 14-2r=0 \\ \Rightarrow r=7 \\ \therefore \mathrm{Required}\mathrm{term}={}^{14}C_7{a}^{14-7}{b}^7=\frac{14!}{(7!{)}^2}{a}^7{b}^7 \end{gathered}$$