Question

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

• A. $2^9.T_{5+1}$
• B. $2^9.T_{7+1}$
• C. $2^9.T_{6+1}$
• D. $2^9.T_{9+1}$

Answer 1

Answer: (C)

$2^9.T_{6+1}$

Since ${(2+3x)}^9=2^9{(1+\frac{3x}{2})}^9$

Now, in the expansion of ${(1+\frac{3x}{2})}^9$, we have:

$\frac{T_{r+1}}{T_r}=\frac{(9-r+1)}{r}\left|\frac{3}{2}x\right|=\frac{(10-r)}{r}|\frac{3}{2}\times \frac{3}{2}|(\because x=3/2)=\left(\frac{10-r}{r}\right)\left(\frac{9}{4}\right)=\frac{90-9r}{4r}\therefore \frac{T_{r+1}}{T_r}\ge 1\Rightarrow \frac{90-9r}{4r}\ge 1\Rightarrow 90\ge 13r\Rightarrow r\le \frac{90}{13}=6\frac{12}{13}\therefore r\le 6\frac{12}{13}$

$\therefore$ Maximum value of $r$ is $6$ So greatest term$=2^9.T_{6+1}$