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

The given expansion is ${(3-5x)}^{15}$, where $x=\frac{1}{5}$

$\begin{array}{l}=3^{15}{\left(1-\frac{5}{3}x\right)}^{15}\\ \frac{T_{r+1}}{T_r}=\frac{n-r+1}{r}\left(\frac{5}{3}x\right)\\ =\frac{15-r+1}{r}\left(\frac{5}{3}\cdot \frac{1}{5}\right)\ge 1\left(\because n=15,x=\frac{1}{5}\right)\\ 16-r\ge 3r\\ \Rightarrow 4r\le 16\\ \Rightarrow r\le 4\\ \therefore r=3,r=4\end{array}$

$\therefore T_4$ and $T_5$ are numerically equal to each other and these are greater than any other term

$\begin{array}{l}{T}_4=3^{15}{}^{15}{C}_3{\left(\frac{-5}{3}\cdot \frac{1}{5}\right)}^3\\ =3^{15}\cdot \frac{15!}{12!3!}{\left(\frac{1}{3}\right)}^3(\therefore \text{\hspace{0.17em}numerical})\\ =3^{15-3}\frac{15\times 14\times 13\times 12!}{12!\times 3\times 2\times 1}\end{array}$

$=3^{12}\left(455\right)$ and 

$T_5=3^{15}{}^{15}{C}_4{\left(\frac{{\displaystyle -5}}{{\displaystyle 3}}\cdot \frac{{\displaystyle 1}}{{\displaystyle 5}}\right)}^4=3^{15}\frac{15!}{\left(11\right)!4!}{\left(\frac{{\displaystyle 1}}{{\displaystyle 3}}\right)}^4$

$=3^{15-4}\frac{15\times 14\times 13\times 12\times 11!}{11!\times 4\times 3\times 2\times 1}=3^{12}\left(455\right)$.