Question

Find numerically the greatest term in the expansion of ${(2x+9)}^9$ when $x=\frac{3}{2}$

Answer 1

Let $T_{r+1}$ be the greatest term in the expansion of ${(2+3x)}^9$ we have

$\frac{T_{r+1}}{T_r}=\left(\frac{9-r+1}{r}\right)\left|\frac{3x}{2}\right|=\left(\frac{10-r}{r}\right)\left|\frac{3x}{2}\right|=\left(\frac{10-r}{r}\right)|\frac{3}{2}\times \frac{3}{2}|=\frac{90-9r}{4r}(\because x=\frac{3}{2})$

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

$\Rightarrow \frac{90-9r}{4r}\ge 1\Rightarrow 90-9r\ge 4r\Rightarrow 90\ge 13r$

$\therefore r\le \frac{90}{13}=6\frac{12}{13}$ or $r=\le 6\frac{12}{13}$

Maximum value of $r$ is $6$

So greatest term $=T_{6+1}={{}^{9}C}_6{\left(2\right)}^{9-6}{\left(3x\right)}^6={{}^{9}C}_3{2}^3{(3\times \frac{3}{2})}^6=\frac{\mathrm{9.8.7}}{\mathrm{1.2.3}}\frac{2^3.3^{12}}{2^6}=\frac{7\times 3^{13}}{2}$

So the greatest term is $\frac{7\times 3^{13}}{2}$