Question

The value of the expression $\sqrt{\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+......\text{upto}99\text{terms}}$ is equal to :

Answer 1

Let, $y=\sqrt{\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+......\text{upto}99\text{terms}}$

or, $y^2=\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+......\text{upto}99\text{terms}$

or, $y^2=\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+......\frac{1}{\sqrt{100}+\sqrt{99}}$

or, $y^2=\frac{\sqrt{2}-1}{(\sqrt{2}{)}^2-(1{)}^2}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}{)}^2-(\sqrt{3}{)}^2}+......\frac{\sqrt{100}-\sqrt{99}}{(\sqrt{100}{)}^2-(\sqrt{99}{)}^2}$

or, $y^2=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+........+(\sqrt{100}-\sqrt{99})$

or, $y^2=\sqrt{100}-1$

or, $y^2=9$

or, $y=\pm 3$

But $y$ can't be $-3$, so $y=3$.