Question

Find the greatest term of the following expansion :
$(1+x{)}^{\frac{21}{2}}$ when $x=\frac{2}{3}$

Answer 1

If $T_{r+1}$ be the greatest term in the expansion of $(1+x{)}^n$ then

$\left|\frac{T_{r+1}}{T_r}\right|\ge 1$

or, $|\frac{n-r+1}{r}.x|\ge 1$

Here, $n=\frac{21}{2}$ and $x=\frac{2}{3}$

or, $|\frac{\frac{21}{2}-r+1}{r}.\frac{2}{3}|\ge 1$

or, $|\frac{(21-2r+2)}{2r}.\frac{2}{3}|\ge 1$

or, $|\frac{(23-2r)}{r}.\frac{1}{3}|\ge 1$

or, $\frac{23-2r}{3r}\ge 1$

or, $23-2r\ge 3r$

or, $5r\le 23$

or, $r\le \frac{23}{5}$

$\therefore r=\left[\frac{23}{5}\right]=4$

$\therefore$ Greatest term is $T_5$ or $5th.$ $[\because T_{r+1}$ is the greatest term$]$