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UPSC CDS II 2024 Math Paper

Question : 8 of 100

Marks: +1, -0

What is

\frac{1}{\sqrt{10}+\sqrt{9}}+\frac{1}{\sqrt{11}+\sqrt{10}}+\frac{1}{\sqrt{12}+\sqrt{11}}+…+\frac{1}{\sqrt{196}+\sqrt{195}}

equal to?

Solution:

Concept:

Rationalize each term to create a telescoping series where most terms cancel out.

Explanation:

For any

n

, we have

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

after rationalization.

Applying this pattern, each term becomes:

\frac{1}{\sqrt{10}+\sqrt{9}} = \sqrt{10} - \sqrt{9}

\frac{1}{\sqrt{11}+\sqrt{10}} = \sqrt{11} - \sqrt{10}

and so on up to

\frac{1}{\sqrt{196}+\sqrt{195}} = \sqrt{196} - \sqrt{195}

Summing all terms, intermediate square roots cancel out, leaving only

-\sqrt{9} + \sqrt{196}

Since

\sqrt{9}=3

and

\sqrt{196}=14

, the sum becomes

-3 + 14 = 11

Answer:

11

(Option C)

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