What is {1}{1 + {2}} + {1}{{2} + {3}} + {1}{{3} + {4}} + ... + {1}{{2020} + {2021}} equal to?

Correct Answer is : 2021−1{2021} - 12021−1

Correct Answer is : 2021−1{2021} - 12021−1

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UPSC CDS 1 2022 Math Paper

Question : 24 of 100

Marks: +1, -0

What is

\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{2020} + \sqrt{2021}}

equal to?

Solution:

Concept:

Rationalizing each term’s denominator creates a telescoping series where all middle terms cancel out.

Explanation:

Take the first term:

\frac{1}{1 + \sqrt{2}}

Multiply numerator and denominator by

\sqrt{2} - 1

(the conjugate):

\frac{1}{1+\sqrt{2}} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2} - 1

Similarly, for any term

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

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

Apply to the entire series from

n=1

n=2020

(\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{2021} - \sqrt{2020})

All intermediate terms cancel, leaving

\sqrt{2021} - 1

Answer:

\sqrt{2021} - 1

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