Question

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

• A. $\frac{\left(7\times 13^3\right)}{2}$
• B. $\frac{\left(7\times 13^2\right)}{2}$
• C. $\frac{\left(7\times 3^{13}\right)}{2}$
• D. None of these

Answer 1

Answer: (C)

We have ${\left(2+3x\right)}^9=2^9{\left(1+\frac{3x}{2}\right)}^9$
Now, $\frac{T_{r+1}}{T_r}=\frac{2^9\left[{}^9{C}_r{\left(\frac{3x}{2}\right)}^r\right]}{2^9\left[{}^9{C}_{r-1}{\left(\frac{3x}{2}\right)}^{r-1}\right]}$
$=\frac{{}^9{C}_r}{{}^9{C}_{r-1}}\left|\frac{3x}{2}\right|=\frac{10-r}{r}\left|\frac{3x}{2}\right|$
$=\frac{10-r}{r}\left(\frac{9}{4}\right)\left[\because x=\frac{3}{2}\right]$
Therefore, $\frac{T_{r+1}}{T_r}\ge 1$
$\Rightarrow \frac{90-9r}{4r}\ge 1$
$\Rightarrow 90-9r\ge 4r$
$\Rightarrow r\le \frac{90}{13}$
$\Rightarrow r\le 6\frac{12}{13}$
Thus, the maximum value of $r$ is $6$. Therefore, the greatest term is $T_{r+1}=T_7$.
$T_7=2^9\left[{}^9{C}_6{\left(\frac{3x}{2}\right)}^6\right]$
$=2^9\cdot {}^9{C}_6{\left(\frac{9}{4}\right)}^6$
$=2^9\cdot \frac{9\times 8\times 7}{3\times 2\times 1}\left(\frac{3^{12}}{2^{12}}\right)=\frac{7\times 3^{13}}{2}$