Question

Find numerically the greatest term in the expansion of ${(3-5x)}^{11}$, when $x=\frac{1}{5}$

Answer 1

Let $T_{r+1}$ be the greatest term in the expansion of $(3-5x)$. We have $\frac{T_{r+1}}{T_r}=\left(\frac{11-r+1}{r}\right)\left|\frac{-5x}{3}\right|$

$=\left(\frac{12-r}{r}\right)\left|\frac{-1}{3}\right|=\frac{12-r}{3r}$ $\because x=\frac{1}{5}$

$\therefore \frac{T_{r+1}}{T_r}\ge 1$

$\Rightarrow \frac{12-r}{3r}\ge 1$

$\Rightarrow 12-r\ge 3r$

$\Rightarrow 12\ge 4r$

$\therefore r\le 3\Rightarrow r=2,3$

So, the greatest terms are $T_{2+1}$ and $T_{3+1}$

$\therefore$ Greatest term (when $r=2$)

$=T_{2+1}={}^{11}{C}_2{3}^9(-5x{)}^2$

$={}^{11}{C}_2{3}^9\cdot 5^2{\left(\frac{1}{5}\right)}^2$

$={}^{11}{C}_2{3}^9=\frac{11.10}{1.2}{3}^9=55\times 3^9$

Greatest term (when $r=3$)

$=T_{3+1}=\left|{}^{11}{C}_3{3}^8\right(-5x{)}^3|$

$=\left|{}^{11}{C}_3{3}^8{(-5\times \frac{1}{5})}^3\right|$ $\because x=\frac{1}{5}$

$={}^{11}{C}_3{3}^8=\frac{\mathrm{11.10.9}}{\mathrm{1.2.3}}{3}^8=55\times 3^9$

Thus, the values of both greatest terms are equal.