Question

Numerically greatest term in the expansion of $(2+3x{)}^9,$ where $x=\frac{3}{2}$ is

• A. $3^{13}\cdot 7$
• B. $\frac{3^{13}\cdot 7}{2}$
• C. $3^{12}\cdot 7$
• D. $\frac{3^{12}\cdot 7}{2}$

Answer 1

The correct option is B $\frac{3^{13}\cdot 7}{2}$

We have, $(2+3x{)}^9=2^9{(1+\frac{3x}{2})}^9$

Now,
$\frac{T_{r+1}}{T_r}=\frac{10-r}{r}\times \frac{3x}{2}=\frac{9(10-r)}{4r}[\because x=\frac{3}{2}]$
For Numerically greatest term,
$\left|\frac{T_{r+1}}{T_r}\right|\ge 1\Rightarrow \left|\frac{9(10-r)}{4r}\right|\ge 1\Rightarrow r\le \frac{90}{13}$
So, Numerically greatest value occurs for $r=6$ which is
$T_{6+1}=2^9\times {}^9{C}_6{\left(\frac{9}{4}\right)}^6=2^9\times {}^9{C}_6{\left(\frac{3}{2}\right)}^{12}=\frac{3^{13}\cdot 7}{2}$