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CDS 2022 · PYQ · Algebra · medium

What is 1/(1+√2) + 1/(√2+√3) + 1/(√3+√4) + ... + 1/(√2020+√2021) equal to?

Answer

The correct answer is B: √2021 - 1. Each term 1/(√n+√(n+1)) = √(n+1) - √n. Telescoping sum = √2021 - 1.

  1. A.√2020 + √2021 - 1
  2. B.√2021 - 1✓ Correct
  3. C.√2020 + 1
  4. D.√2021 + 1

Explanation

Each term 1/(√n+√(n+1)) = √(n+1) - √n. Telescoping sum = √2021 - 1.
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