Question

Prove that $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{5}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}$

$=2$

Answer 1

We will take general term and we will solve it

General term =$\frac{1}{\sqrt{n}+\sqrt{n+1}}$

$\frac{1}{\sqrt{n}+\sqrt{n+1}}\times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}$

$=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}$

$=\sqrt{n+1}-\sqrt{n}$

Given question becomes ;

$\sqrt{2}-1+\sqrt{3}-\sqrt{2}+......+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8}$

$\Rightarrow \sqrt{9}-1=2$