Question

The numerically greatest term in the expansion of $(3+2x{)}^{14}$ when $x=4/5$ is

• A. 4th term
• B. 5th term
• C. 6th term
• D. 7th term

Answer 1

Answer: (C)

6th term

We know that in expansion of $(x+y{)}^n$
$\frac{T_{r+1}}{T_r}=\frac{(n-r+1)}{r}\frac{y}{x}$
Let $T_{r+1}$ be the greatest term
$|\frac{T_{r+1}}{T_r}|=\frac{14-r+1}{r}|\frac{2x}{3}|$
$|\frac{T_{r+1}}{T_r}|=\frac{15-r}{r}|\frac{8}{15}|$
$\Rightarrow |\frac{T_{r+1}}{T_r}|=\frac{120-8r}{15r}$
For greatest term,
$|\frac{T_{r+1}}{T_r}|\ge 1$
$\Rightarrow \frac{120-8r}{15r}\ge 1$
$\Rightarrow 23r\le 120$
$\Rightarrow r=5$
So, the greatest term is $T_{r+1}$
$T_{5+1}{=}^{14}{C}_5\left(3{)}^9\right(2x{)}^5$