Question

When $x=9,y=4$, value of the numerically greatest term in the expansion of $(2x-3y{)}^{28}$ is

• A. ${}^{28}{C}_{12}{.2}^{40}{.3}^{44}$
• B. ${}^{-28}{C}_{12}{.2}^{40}{.3}^{44}$
• C. ${}^{28}{C}_{11}{.2}^{39}{.3}^{45}$
• D. ${}^{-28}{C}_{11}{.2}^{39}{.3}^{45}$

Answer 1

The correct option is D ${}^{-28}{C}_{11}{.2}^{39}{.3}^{45}$

$(2x-3y{)}^{28}$
For $x=9$ and $y=4$
$(18-12{)}^{28}$
$=\left(18{)}^{28}\right(1-\frac{12}{18}{)}^{28}$
$=\left(18{)}^{28}\right(1-\frac{2}{3}{)}^{28}$
Therefore numerically greatest term will be given by.
$\frac{29\left(\frac{2}{3}\right)}{1+\frac{2}{3}}$
$=\frac{29\left(2\right)}{5}$
$=11.6$
Hence $r=11$
Hence option D is the correct option.