Question

Middle term in the expansion of (a³ + ba)²⁸ is ___________.

Answer 1

In (a³ + ba)²⁸
Here n = 28
So, there is one middle term
$$\begin{gathered} \mathrm{i}.\mathrm{e}.\left(\frac{28}{2}+1\right)\mathrm{th}\mathrm{term}\mathrm{which}\mathrm{is}15\mathrm{th}\mathrm{term} \\ \therefore \mathrm{Middle}\mathrm{term}\mathrm{is}{T}_{15} \end{gathered}$$

$\begin{array}{rcl}& & \mathrm{i}.\mathrm{e}.T_{14+1}{=}^{28}{C}_{14}{\left(a^3\right)}^{28-14}{\left(ba\right)}^{14}\left(\mathrm{using}{T}_{r+1}{=}^n{C}_r{x}^r{y}^{n-r}\mathrm{for}{(x+y)}^n\right)\\ & & {=}^{28}{C}_{14}{\left(a^3\right)}^{14}{b}^{14}{a}^{14}\\ & =& {}^{28}{C}_{14}{a}^{56}{a}^{14}\end{array}$