Examveda
Examveda

Determined the value of $$\frac{1}{{\sqrt 1 + \sqrt 2 }}{\text{ + }}$$  $$\frac{1}{{\sqrt 2 + \sqrt 3 }}\, + $$   $$\frac{1}{{\sqrt 3 + \sqrt 4 }}\, + $$   $$...... + $$   $$\frac{1}{{\sqrt {120} + \sqrt {121} }}{\text{ = ?}}$$

A. 8

B. 10

C. $$\sqrt {120} $$

D. $$12\sqrt 2 $$

Answer: Option B

Solution(By Examveda Team)

Given expressing,
$$ = \frac{1}{{\sqrt 1 + \sqrt 2 }}{\text{ + }}\frac{1}{{\sqrt 2 + \sqrt 3 }}$$     $$ + \frac{1}{{\sqrt 3 + \sqrt 4 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {120} + \sqrt {121} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }}{\text{ + }}\frac{1}{{\sqrt 3 + \sqrt 2 }}$$     $$ + \frac{1}{{\sqrt 4 + \sqrt 3 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }} \times $$   $$\frac{{\sqrt 2 - \sqrt 1 }}{{\sqrt 2 - \sqrt 1 }}{\text{ + }}$$   $$\frac{1}{{\sqrt 3 + \sqrt 2 }} \times $$   $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} + $$   $$\frac{1}{{\sqrt 4 + \sqrt 3 }} \times $$   $$\frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} + $$   $$...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }} \times $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{\sqrt {121} - \sqrt {120} }}$$
$$ = \frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}$$     $$ + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}}$$   $$ + ...... + $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{121 - 120}}$$
$$ = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 $$     $$ + \sqrt 4 - \sqrt 3 $$   $$ + ...... + $$   $$\sqrt {121} - \sqrt {120} $$
$$ = - 1 + \sqrt {121} $$
$$ = - 1 + 11$$
$$ = 10$$

This Question Belongs to Arithmetic Ability >> Square Root And Cube Root

Join The Discussion

Related Questions on Square Root and Cube Root