Question

The value of :
$\frac{1}{\sqrt{9}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{1}}$ is :

• A. $7$
• B. $5$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

$2$

Let us take general term : $\frac{1}{\sqrt{n+1}+\sqrt{n}}$

Multiply and divide by its complement

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

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

$\therefore \frac{1}{\sqrt{9}+\sqrt{8}}=\sqrt{9}-\sqrt{8}$

Similarly we will simplify the terms

Question becomes :

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

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

$\Rightarrow 3-1=2$

Answer is 2