Question

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

• A. $\frac{\left(7\times13^{3}\right)}{2}$
• B. $\frac{\left(7\times13^{2}\right)}{2}$
• C. $\frac{\left(7\times3^{13}\right)}{2}$
• 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} $
$= 2^{9}\cdot \frac{9\times8\times 7}{3\times 2\times 1}\left(\frac{3^{12}}{2^{12}}\right) = \frac{7\times 3^{13}}{2}$