Question

Q.7 in the attachment

Answer 1

Answer:

3

Step-by-step explanation:

Given :

To find the value of the expression :

$\sqrt{\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}....upto \ 99 \ terms \ is \ equal \ to}$

Solution :

The fractions are in the form :

⇒ $\frac{1}{\sqrt{n+1}+\sqrt{n}}$

By taking conjugate,

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

⇒ $\frac{\sqrt{n+1} - \sqrt{n} }{(\sqrt{n+1})^2-(\sqrt{n})^2} =\frac{\sqrt{n+1} - \sqrt{n} }{n+1-n}=\frac{\sqrt{n+1} - \sqrt{n} }{1}=\sqrt{n+1} - \sqrt{n}$

∴ $\frac{1}{\sqrt{2}+\sqrt{1}} = \sqrt{2} - \sqrt{1}$

∴ $\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3} -\sqrt{2}$

∴ $\frac{1}{\sqrt{4}+\sqrt{3}}=\sqrt{4} -\sqrt{3}$

$\sqrt{\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}...}$

⇒ $\sqrt{(\sqrt{2} -\sqrt{1})+(\sqrt{3} - \sqrt{2}) + (\sqrt{4}- \sqrt{3})+... +(\sqrt{99}-\sqrt{98})+ (\sqrt{100} -\sqrt{99})}$

⇒ $\sqrt{\sqrt{100} -\sqrt{1}} = \sqrt{10 - 1} = \sqrt{9} = 3$