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UPSC CDS 2 2020 Math Paper

Question : 12 of 100

Marks: +1, -0

\frac{1}{1+\sqrt{2}}+ \frac{1}{\sqrt{2} + \sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}?

Solution:

Concept:

Rationalize each denominator to create a telescoping series.

Explanation:

The given sum is

S = \sum_{n=1}^{99} \frac{1}{\sqrt{n}+\sqrt{n+1}}

Rationalize each term by multiplying numerator and denominator by the conjugate

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

For any

n

\frac{1}{\sqrt{n}+\sqrt{n+1}} \times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}} = \frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n} = \sqrt{n+1}-\sqrt{n}

Thus

S = (\sqrt{2}-\sqrt{1}) + (\sqrt{3}-\sqrt{2}) + (\sqrt{4}-\sqrt{3}) + \dots + (\sqrt{100}-\sqrt{99})

All intermediate terms cancel, leaving

S = \sqrt{100} - \sqrt{1} = 10 - 1 = 9

Answer:

9

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