Question

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

• A. ${}^9{C}_6{.2}^9(3/2{)}^{12}$
• B. ${}^9{C}_6{.2}^6.(3/2{)}^6$
• C. ${}^9{C}_5{.2}^9.(3/2{)}^{10}$
• D. ${}^9{C}_4{.2}^9.(3/2{)}^8$

Answer 1

The correct option is B ${}^9{C}_6{.2}^6.(3/2{)}^6$

Let $T_r$ and $T_{r+1}$ denote the $r^{th}$ and ${(r+1)}^{th}$ terms in the expansion of ${(2+3x)}^9$

then, $T_r{=}_{r-1}^{9}C{\left(2\right)}^{9-r+1}{\left(3x\right)}^{r-1}{T}_{r+1}{=}_r^{9}C{\left(2\right)}^{9-r}{\left(3x\right)}^{r}so,\frac{T_{r+1}}{T_r}=\frac{{}^{9}C{\left(2\right)}^{9-r}{\left(3x\right)}^r}{{}_{r-1}^{9}C{\left(2\right)}^{9-r+1}{\left(3x\right)}^{r-1}}=\frac{9-r}{r}\left(\frac{3}{2}x\right)when,x=\frac{3}{2}\frac{T_{r+1}}{T_r}=\frac{81-9r}{4r}now,\frac{T_{r+1}}{T_r}>1\Rightarrow \frac{81-9r}{4r}>1\Rightarrow r<6\frac{3}{13}$

Thus ${(6+1)}^{th}=7^{th}$ term is the greatest term

$\therefore T_7{=}_6^{9}C{\left(2\right)}^{9-6}{\left(3.\frac{3}{2}\right)}^3{=}_6^{9}C{2}^6.{\left(\frac{3}{2}\right)}^6$