Question

Find the numerically greatest term in the expansion of ${(3-5x)}^{15}$ when $x=\frac{1}{5}.$

Answer 1

Let $r^{th}$ and ${(r+1)}^{th}$ be two consecutive terms in the expansion of ${(3-5x)}^{15}$
$T_{r+1}\ge T_r$
${}^{15}{\text{C}}_r{3}^{15-r}{\left(|-5x|\right)}^r{\ge }^{15}{\text{C}}_{r-1}{3}^{15-(r-1)}{\left(|-5x|\right)}^{r-1}$
$\frac{\left(15\right)!}{(15-r)!r!}|-5x|\ge \frac{3.\left(15\right)!}{(16-r)!(r-1)!}$
$5.\frac{1}{5}(16-r)\ge 3r$
$16-r\ge 3r$
$4r\le 16$
$r\le 4$

Hence, $4$th term is the greatest.