Question

Find the greatest term of the following expansion :
$(1+x{)}^{-7}$ when $x=\frac{4}{15}$

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, $n=-7$ and $x=\frac{4}{15}$

or, $|\frac{-7-r+1}{r}.\frac{4}{15}|\ge 1$

or, $|\frac{(-6-r)}{r}.\frac{4}{15}|\ge 1$

or, $|\frac{(6+r)}{r}.\frac{4}{15}|\ge 1$

or, $\frac{24+4r}{15r}\ge 1$

or, $24+4r\ge 15r$

or, $11r\le 24$

or, $r\le \frac{24}{11}$

$\therefore r=\left[\frac{24}{11}\right]=2$

$\therefore$ Greatest term is $T_3$ or $3rd.$