Question

The numerically greatest term in the expansion of $(3-2x{)}^9$ when $x=1$ is

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

Answer 1

The correct option is B 5th term

$=(3-2x{)}^9$

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

$=3^9(1-\frac{2}{3}{)}^9$

let Trt, be the greatest term this exp.

Tr+1 $\ge Tr$

${}^9{C}_r{\left(\frac{-2}{3}\right)}^r{\ge }^9{C}_{r-1}{\left(\frac{-2}{3}\right)}^{r-1}$

$\Rightarrow \frac{9!}{r!|9-r|!}{\left(\frac{-2}{3}\right)}^{r-r+1}\ge \frac{9!}{(r-1)!|9-r+1|!}$

$\Rightarrow \frac{1}{r|r-1|!|9-r|!}\left|\frac{-2}{3}\right|\ge \frac{1}{(r-1)!(10-r)(9-r)!}$

$=(-2(10-r\left)\right)\ge \left(3r\right)$

$=20-2r\ge 3r$

$=20\ge 5r$

$=r\le 4$

$T_5$ is the greatest term