Question

Find the greatest term of the following expansion :
$(1-7x{)}^{-\frac{11}{4}}$ when $x=\frac{1}{8}$

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, $X=-7x,$ $n=-\frac{11}{4}$ and $x=\frac{1}{8}$

or, $|\frac{-\frac{11}{4}-r+1}{r}.\frac{(-7)}{8}|\ge 1$

or, $|\frac{(-11-4r+4)}{4r}.\frac{(-7)}{8}|\ge 1$

or, $|\frac{(-7-4r)}{4r}.\frac{(-7)}{8}|\ge 1$

or, $|\frac{(7+4r)}{4r}.\frac{7}{8}|\ge 1$

or, $\frac{49+28r}{32r}\ge 1$

or, $49+28r\ge 32r$

or, $4r\le 49$

or, $r\le \frac{49}{4}$

$\therefore r=\left[\frac{49}{4}\right]=12$

$\therefore$ Greatest term is $T_{13}$ or $13th.$ $[\because T_{r+1}$ is the greatest term$]$