Question

If $S=\frac{1}{\sqrt{2}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{11}}+\dots \dots n\text{terms}$, then which of the following is/are correct?

• A. $S=\frac{\sqrt{3n+2}-\sqrt{2}}{3}$
• B. $S=\frac{n}{\sqrt{2+3n}-\sqrt{2}}$
• C. $S<n$ always
• D. $S=\frac{n}{2}$ has $1$ solution

Answer 1

Answer: (A), (C)

The correct options are
A $S=\frac{\sqrt{3n+2}-\sqrt{2}}{3}$
C $S<n$ always

Given : $S=\frac{1}{\sqrt{2}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{11}}+\dots \dots n\text{terms}$
$\Rightarrow S=\frac{1}{3}[\sqrt{5}-\sqrt{2}+\sqrt{8}-\sqrt{5}+\dots +\sqrt{3n+2}-\sqrt{3n-1}]\Rightarrow S=\frac{(\sqrt{3n+2}-\sqrt{2})}{3}\Rightarrow S=\frac{n}{(\sqrt{3n+2}+\sqrt{2})}$

Since, $\sqrt{3n+2}>n$
$\therefore S=\frac{n}{\sqrt{2+3n}+\sqrt{2}}<n$

Now,
$S=\frac{n}{2}\Rightarrow \sqrt{2+3n}+\sqrt{2}=2\Rightarrow 2+3n=(2-\sqrt{2}{)}^2\Rightarrow n=\frac{4-4\sqrt{2}}{3}$
As $n$ is number of terms, so it has to be integer, therefore $S=\frac{n}{2}$ has no solution.