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,$$ = \frac{1}{{\sqrt 1 + \sqrt 2 }}{\text{ + }}\frac{1}{{\sqrt 2 + \sqrt 3 }}$$ $$ + \frac{1}{{\sqrt 3 + \sqrt 4 }}$$ $$ + ...... + $$ $$\frac{1}{{\sqrt {120} + \sqrt {121} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }}{\text{ + }}\frac{1}{{\sqrt 3 + \sqrt 2 }}$$ $$ + \frac{1}{{\sqrt 4 + \sqrt 3 }}$$ $$ + ...... + $$ $$\frac{1}{{\sqrt {121} + \sqrt {120} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }} \times $$ $$\frac{{\sqrt 2 - \sqrt 1 }}{{\sqrt 2 - \sqrt 1 }}{\text{ + }}$$ $$\frac{1}{{\sqrt 3 + \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} + $$ $$\frac{1}{{\sqrt 4 + \sqrt 3 }} \times $$ $$\frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} + $$ $$...... + $$ $$\frac{1}{{\sqrt {121} + \sqrt {120} }} \times $$ $$\frac{{\sqrt {121} - \sqrt {120} }}{{\sqrt {121} - \sqrt {120} }}$$
$$ = \frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}$$ $$ + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}}$$ $$ + ...... + $$ $$\frac{{\sqrt {121} - \sqrt {120} }}{{121 - 120}}$$
$$ = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 $$ $$ + \sqrt 4 - \sqrt 3 $$ $$ + ...... + $$ $$\sqrt {121} - \sqrt {120} $$
$$ = - 1 + \sqrt {121} $$
$$ = - 1 + 11$$
$$ = 10$$
Related Questions on Square Root and Cube Root
The least perfect square, which is divisible by each of 21, 36 and 66 is:
A. 213444
B. 214344
C. 214434
D. 231444
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