Question

$\frac{1}{\sqrt{2}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{11}}+......n$ terms value of the above expression is

• A. $\frac{\sqrt{3n+2}-\sqrt{2}}{3}$
• B. $\frac{n}{\sqrt{3n+2}+\sqrt{2}}$
• C. less than n
• D. less than $\sqrt{\frac{n}{3}}$

Answer 1

Answer: (A), (B), (C)

$\frac{n}{\sqrt{3n+2}+\sqrt{2}}$
less than n
$\frac{\sqrt{3n+2}-\sqrt{2}}{3}$

The value of $\frac{1}{\sqrt{2}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{1}1}$+ ........ n terms
$=\frac{\sqrt{5}-\sqrt{2}}{3}+\frac{\sqrt{8}-\sqrt{5}}{3}+\frac{\sqrt{1}1-\sqrt{8}}{3}+....+\frac{\sqrt{5+(n-1)3}-\sqrt{2+(n-1)3}}{3}$
$=\frac{\sqrt{3n+2}-\sqrt{2}}{3}=\frac{3n+2-2}{3(\sqrt{3n+2}+\sqrt{2})}=\frac{n}{\sqrt{3n+2}+\sqrt{2}}$
$=\frac{n}{\sqrt{2+3n}+\sqrt{2}}<\frac{n}{\sqrt{3n}}<n$