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JEE MainMathematicsBinomial Theorem

The remainder when the sum of the series 3^1 + 3^2 + 3^3 + + 3⁹⁸ is divided by 13 is

Options

  1. A11
  2. B9
  3. C4
  4. D12

Correct answer

D. 12

Step-by-step solution

Let S = 3^1 + 3^2 + 3^3 + + 3⁹⁸ This is a geometric progression with first term a = 3 , common ratio r = 3 , and number of terms n = 98 . S = 3(3⁹⁸ - 1) 3 - 1 = 3⁹⁹ - 3 2 We need to find the remainder when S is divided by 13 . Since S is an integer obtained by dividing by 2 , we first find the remainder of the numerator 3⁹⁹ - 3 when divided by 26 (which is 13 2 ). 3⁹⁹ = (3^3)³³ = 27³³ = (26 + 1)³³ Using the Binomial Theorem: (26 + 1)³³ = ³³C₀ 26³³ + ³³C₁ 26³² + + ³³C₃₃ 1³³ = 26k + 1 for some integer k . So, 3⁹⁹ - 3

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