First slide

Binomial theorem for positive integral Index

Question

The greatest term in the expansion of $\sqrt{5} {(1 + \frac{1}{\sqrt{5}})}^{20}$
is :

Moderate

A

$^{20} C_{5} (\frac{1}{25})$
B

$^{20} C_{5} (\frac{1}{5})$
C

$^{20} C_{5}$
D

$^{20} C_{5} (5)$

Solution

We have
$T_{r + 1} =^{20} C_{r} \frac{\sqrt{5}}{(\sqrt{5})^{r}} =^{20} C_{r} \frac{1}{(\sqrt{5})^{r - 1}}$
Suppose $(r + 1)$ th term is the greatest term. Then
$\begin{array}{l} \frac{T_{r + 1}}{T_{r}} > 1 and \frac{T_{r + 1}}{T_{r + 2}} > 1 \\ \Rightarrow \frac{20 - r}{r + 1} \frac{1}{\sqrt{5}} > 1 and \frac{r + 2}{19 - r} \sqrt{5} > 1 \\ \Rightarrow 20 - r > \sqrt{5} r + \sqrt{5} and r \sqrt{5} + 2 \sqrt{5} > 19 - r \\ \Rightarrow 19 - 2 \sqrt{5} < (\sqrt{5} + 1) r < 20 - \sqrt{5} \\ \Rightarrow \frac{1}{4} (19 - 2 \sqrt{5}) (\sqrt{5} - 1) < r < \frac{1}{4} (20 - \sqrt{5}) (\sqrt{5} - 1) \\ \Rightarrow \frac{1}{4} (21 \sqrt{5} - 29) < r < \frac{1}{4} (21 \sqrt{5} - 25) \end{array}$
$\Rightarrow 4.48 < r < 5.49 \Rightarrow r = 5,$
$∴ greatest term =^{20} C_{5} {(\frac{1}{\sqrt{5}})}^{4} =^{20} C_{5} (\frac{1}{25})$

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