Question

Value of the numerically greatest term in $\sqrt{5}{[1+\frac{1}{\sqrt{5}}]}^{20}$

Answer 1

Let $T_{r+1}$ be numerically greatest term

$\therefore$ $\frac{T_{r+1}}{T_r}>1$

$\Rightarrow \frac{n_{C_r}{\left(\frac{1}{\sqrt{5}}\right)}^{20-r}}{n_{C_{r-1}}{\left(\frac{1}{\sqrt{5}}\right)}^{20-(r-1)}}>1$

$\Rightarrow {20}_{C_r}{\left(\frac{1}{\sqrt{5}}\right)}^{20}\times {\left(\sqrt{5}\right)}^r>{20}_{C_{r-1}}{\left(\frac{1}{\sqrt{5}}\right)}^{20}\times \frac{{\left(\sqrt{5}\right)}^r}{\sqrt{5}}$

$\Rightarrow \frac{{20}_{C_r}}{{20}_{C_{r-1}}}>\frac{1}{\sqrt{5}}$

$\Rightarrow \frac{\frac{20!}{(20-r)!r!}}{\frac{20!}{(20-r)!(r-1)!}}>\frac{1}{\sqrt{5}}$

$\Rightarrow \frac{(21-r)}{r}>\frac{1}{\sqrt{5}}$

$\Rightarrow 21\sqrt{5}-r\sqrt{5}>r$

$\Rightarrow 21\sqrt{5}>(\sqrt{5}+1)r$

$\Rightarrow r\le \frac{21\sqrt{5}}{\sqrt{5}+1}=14.51$ $\therefore$ $r=14$

$\therefore$ numerically greatest term is $T_{15}={20}_{C_{14}}{\left(\frac{1}{\sqrt{5}}\right)}^6.\sqrt{5}$

$={20}_{C_r}{\left(\frac{1}{\sqrt{5}}\right)}^5$

$=693.3599$