$\text{ Which term is greatest (Numerically) in the expansion }{\left(1+\frac{3x}{5}\right)}^{15}\text{ when }x=\frac{1}{3}$

$\text{ We have }n=15;X=\frac{3x}{5}=\frac{3}{5}·\frac{1}{3}\left(\because x=\frac{1}{3}\right)=\frac{1}{5}$

$\text{ Now }\frac{(n+1)\left|X\right|}{\left|X\right|+1}=\frac{(15+1)\times \frac{1}{5}}{\frac{1}{5}+1}=\frac{16\times \frac{1}{5}}{\frac{6}{5}}=\frac{18}{6}=2.6=2+0.6=p+F$

$\therefore \text{ Numerically greatest term is }{T}_{p+1}=T_{2+1}=T_3$