Question

Find the value of greatest term in the expansion of $\sqrt{3}{\left[1+\frac{1}{\sqrt{3}}\right]}^{20}$ .

Answer 1

Consider ${(1+\frac{1}{\sqrt{3}})}^{20}$

If $T_{r+1}$ is numerically greatest

term, then $\frac{T_{r+1}}{T_r}>1$

or $\frac{n-r+1}{r}\times x>1$

where $n=20$, $x=\frac{1}{\sqrt{3}}$, $r=?$

$\therefore \left(\frac{20-r+1}{r}\right)\times \frac{1}{\sqrt{3}}>1$

or $20-r+1>\sqrt{3}r$

or $\sqrt{3}r+r<21$

or $r<\frac{21}{\sqrt{3}+1}$

or $r<7.68$ $\to \left(1\right)$

Similarly $\frac{T_{r+1}}{T_{r+2}}>1$

or $\frac{r+1}{r-1}\times \frac{1}{x}>1$

or $\frac{r+1}{20-r}>\frac{1}{\sqrt{3}}$

or $\sqrt{3r}+\sqrt{3}>20-r$

or $r>\frac{20-\sqrt{3}}{\sqrt{3}+1}$

or $r>6.68\to \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

only integer value of $r=7$

$\therefore$ greatest term $=\sqrt{3}$ ${}^{20}{C}_7{\sqrt{3}}^{13}$

$={}^{20}{C}_7(\sqrt{3}{)}^{14}={}^{20}{C}_7\times 3^7$.