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JEE Main202125 Jul 2021Evening ShiftMathematicsBinomial TheoremActual

The lowest integer which is greater than 1 + 1 10 100 10 100 is

Options

  1. A3
  2. B4
  3. C2
  4. D1

Correct answer

A. 3

Step-by-step solution

Consider P = 1 + 1 10 100 10 100 , Let x = 10 100 ⇒ P = 1 + 1 x x ⇒ P = 1 + x 1 x + x x - 1 ⌊ 2 · 1 x 2 + x x - 1 x - 2 3 3 · 1 x 3 + … (upto 10 100 + 1 terms) ⇒ P = 1 + 1 + 1 ⌊ 2 - 1 2 x + 1 ⌊ 3 - … + … so on ⇒ P = 2 + (Positive value less than 1 ⌊ 2 + 1 ⌊ 3 + 1 ⌊ 4 + … ) Also e = 1 + 1 ⌊ 1 + 1 ⌊ 2 + 1 ⌊ 3 + 1 ⌊ 4 + … ⇒ 1 ⌊ 2 + 1 ⌊ 3 + 1 ⌊ 4 + … = e - 2 ⇒ P

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