Question

If the sum of the binomial coefficients of the expansion ${\left(2x+\frac{1}{x}\right)}^n$ is equal to 256, then the term independent of x is
(a) 1120
(b) 1020
(c) 512
(d) none of these

Answer 1

(a) 1120

$$\begin{gathered} \mathrm{Suppose}(r+1)\mathrm{th}\mathrm{tem}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}\mathrm{is}\mathrm{independent}\mathrm{of}x\mathit{.} \\ \mathrm{Then},\mathrm{we}\mathrm{have} \\ {T}_{r+1}={}^{n}C_r(2x{)}^{n-r}{\left(\frac{1}{x}\right)}^r \\ ={}^{n}C_r{2}^{n-r}{x}^{n-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have} \\ n-2r=0 \\ \Rightarrow r=n/2 \\ \therefore \mathrm{Required}\mathrm{term}={}^{n}C_{n/2}{2}^{n-n/2}=\frac{n!}{{\left[\left(n/2\right)!\right]}^2}{2}^{n/2} \\ \mathrm{We}\mathrm{know}: \\ \mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{expansion}=256 \\ \mathrm{Thus},\mathrm{we}\mathrm{have} \\ {2}^n.1^n=256 \\ \Rightarrow n=8 \\ \therefore \mathrm{Required}\mathrm{term}=\frac{8!}{\left(4\right)!\left(4\right)!}{2}^4=1120 \end{gathered}$$