Question

If r^th term is the middle term in the expansion of ${\left(x^2-\frac{1}{2x}\right)}^{20}$, then ${\left(r+3\right)}^{th}$ term is
(a) ${}^{20}C_{14}\left(\frac{x}{2^{14}}\right)$

(b) ${}^{20}C_{12}{x}^2{2}^{-12}$

(c) $-{}^{20}C_{7}x,2^{-13}$

(d) none of these

Answer 1

(c) $-{}^{20}C_{7}x\times 2^{-13}$
Here n is even
So, The middle term in the given expansion is $\left(\frac{20}{2}+1\right)\mathrm{th}=11\mathrm{th}\mathrm{term}$
Therefore, (r + 3)th term is the 14th term.

$$\begin{gathered} T_{14}={}^{20}C_{13}(x^2{)}^{20-13}{\left(\frac{-1}{2x}\right)}^{13} \\ ={\left(-1\right)}^{13}{}^{20}C_{13}\frac{x^{14-13}}{2^{13}} \\ =-{}^{20}C_{7}x{2}^{-13} \end{gathered}$$