Question

In the expansion of $(\frac{1}{2}+\frac{2x}{3})$, where $x=-\frac{1}{2}$, it is known that $3^{rd}$ term is the greatest term. Find the possible integral values of $n$.

Answer 1

Given : ${(\frac{1}{2}+\frac{2x}{3})}^x;x=-\frac{1}{2}$ $;T_3$ is greatest term.

To find : Possible integral values of $n.$

Sol. $:T_3\ge T_4;$ $T_3\ge T_2$

Case $1:\to$ $T_3\ge T_4$

$={}^n{C}_2{\left(\frac{1}{2}\right)}^{n-2}.{\left(\frac{2x}{3}\right)}^2\ge {}^n{C}_3{\left(\frac{1}{2}\right)}^{n-3}.{\left(\frac{2x}{3}\right)}^3$

$=\frac{n-2}{3}\times \frac{2x}{3}\le \frac{1}{2}$

$\Rightarrow$ $n=-\frac{1}{2}=\frac{n-2}{3}\times \frac{-1}{3}\le \frac{1}{2}$

$\Rightarrow$ $n\ge \frac{-5}{2}$

Case $2:\to$ $T_3\ge T_2$

$={}^n{C}_2{\left(\frac{1}{2}\right)}^{n-2}.{\left(\frac{2x}{3}\right)}^2\ge {}^n{C}_1{\left(\frac{1}{2}\right)}^{n-1}.{\left(\frac{2x}{3}\right)}^1$

$\Rightarrow$ $=(n-1)\times \frac{2x}{3}\ge 1$ $;n=\frac{-1}{2}$

$\Rightarrow$ $=(n-1)\times \frac{-1}{3}\ge 1$

$\Rightarrow$ $=n-1\le -3$

$\Rightarrow$ $=n\le -2$

$\therefore$ $\frac{-5}{2}\le n\le -2$

Hence, integral value of $n=-2.$