Question

Find the value of the greatest term in the expansion of $\sqrt{3}{(1+\frac{1}{\sqrt{3}})}^{20}$.

• A. $7$
• B. $8$
• C. $9$
• D. $6$

Answer 1

The correct option is B $8$

Let $T_{r+1}$ be the greatest term, then $T_r<T_{r+1}>T_{r+2}$
Consider : $T_{r+1}>T_r$
${\Rightarrow }^{20}{C}_r\left(\frac{1}{\sqrt{3}}{)}^r{>}^{20}{C}_{r-1}\right(\frac{1}{\sqrt{3}}{)}^{r-1}$
$\Rightarrow \frac{\left(20\right)!}{(20-r)!r!}(\frac{1}{\sqrt{3}}{)}^r>\frac{\left(20\right)!}{(21-r)!(r-1)!}(\frac{1}{\sqrt{3}}{)}^{r-1}$
$\Rightarrow r<\frac{21}{\sqrt{3}+1}$
$\Rightarrow r<7.686$ ......... (i)
Similarly, considering $T_{r+1}>T_{r+2}$
$\Rightarrow r>6.69$ .......... (ii)
From (i) and (ii), we get
$r=7$
Hence greatest term = $T_8$ = $\frac{25840}{9}$