Q. The largest term in the expansion of ${(3 + 2x)}^{50}$, where $x = \frac{1}{5}$ is

A) 5th
B) 3rd
C) 7th
D) 6th
Explanation

(c)
${\because(3 + 2x)}^{50} = 3^{50}\left( 1 + \frac{2x}{3} \right)^{50}$
Here, $T_{r + 1} = 3^{50}\ ^{50}C_{r}\left( \frac{2x}{3} \right)^{r}$
and $T_{r} = 3^{50}\ ^{50}C_{r - 1}\left( \frac{2x}{3} \right)^{r - 1}$
But $x = \frac{1}{5}$
$\therefore\frac{T_{r + 1}}{T_{r}} \geq 1 \Rightarrow \frac{\ ^{50}C_{r}}{\ ^{50}C_{r - 1}}\frac{2}{3} \bullet \frac{1}{5} \geq 1$
$\Rightarrow \ 102 - 2r \geq 15r\ \Rightarrow \ r \leq 6$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.