Question

The largest term in the expansion of $(3+2x{)}^{50}$, where $x=\frac{1}{5}$, is
$I$. $5^{\text{th }}$
$II$. $3^{\text{rd }}$
$III$. $7^{\text{th }}$
$IV$. $6^{\text{th }}$
Choose the correct option

• A. Only I
• B. Only II
• C. Both I and IV
• D. Both III and IV

Answer 1

Answer: (D)

$\because (3+2x{)}^{50}=3^{50}{\left(1+\frac{2x}{3}\right)}^{30}$
Here, $T_{r+1}=3^{50}{}^{50}{C}_r{\left(\frac{2x}{3}\right)}^{r-1}$
and $T_{r+1}=3^{50}{}^{50}{C}_r{\left(\frac{2x}{3}\right)}^{r-1}$
But $x=\frac{1}{5}$[given]
$\therefore \frac{T_{r+1}}{T_r}\ge 1$
$\Rightarrow \frac{{}^{50}{C}_r}{{}^{50}{C}_{r-1}}\frac{2}{3}\cdot \frac{1}{5}\ge 1$
$\Rightarrow 102-2r\ge 15r$
$\Rightarrow r\le 6$
$\Rightarrow r=6$
Therefore, there are two greatest terms $T_r$ and $T_{r+1}$ i.e., $T_6$ and $T_7$