Question

Find the sum of the expression

\(\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{ {\sqrt2 + \sqrt 3 } }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + \ldots + \frac{1}{{\sqrt {80} + \sqrt {81} }}\)

1. 7
2. 8
3. 9
4. 10

Answer 1

Correct Answer - Option 2 : 8

Calculation:

Given:

\(\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + \ldots + \frac{1}{{\sqrt {80} + \sqrt {81} }}\)

We can write as,

\( = \;\frac{1}{{\sqrt 2 + \sqrt 1 }} \times \left( {\frac{{\sqrt 2 - \sqrt 1 }}{{\sqrt 2 - \sqrt 1 }}} \right) + \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \left( {\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}} \right) + \frac{1}{{\sqrt 4 + \sqrt 3 }} \times \left( {\frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }}} \right) + \ldots + \frac{1}{{\sqrt {81} + \sqrt {80} }} \times \left( {\frac{{\sqrt {81} - \sqrt {80} }}{{\sqrt {81} - \sqrt {80} }}} \right)\)

∵  a² – b² = (a + b) × (a – b)

\( = \;\frac{{\sqrt 2 - \sqrt 1 }}{{{{\left( {\sqrt 2 } \right)}^2} - {{\left( {\sqrt 1 } \right)}^2}}} + \frac{{\sqrt 3 - \sqrt 2 }}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} + \frac{{\sqrt 4 - \sqrt 3 }}{{{{\left( {\sqrt 4 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}} + \ldots + \frac{{\sqrt {81} - \sqrt {80} }}{{{{\left( {\sqrt {81} } \right)}^2} - {{\left( {\sqrt {80} } \right)}^2}}}\)

\( = \;\frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}} + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}} + \ldots + \frac{{\sqrt {81} - \sqrt {80} }}{{81 - 80}}\)

\( = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 + \sqrt 4 - \sqrt 3 + \ldots \sqrt {81} - \sqrt {80} \)

After cancellation of suitable terms,

We get,

\( = \sqrt {81} - \sqrt 1 \Rightarrow 9 - 1 = 8\)