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KVPY2020MathematicsSequences and Series

Let a = ∑ n = 101 200 2 n ∑ k = 101 n 1 k ! and b = ∑ n = 101 200 2 201 - 2 n n ! Then a b is

Options

  1. A1
  2. B3 2
  3. C2
  4. D5 2

Correct answer

A. 1

Step-by-step solution

a = ∑ n = 101 200 2 n ∑ k = 101 n 1 k ! = 2 101 101 ! + 2 102 1 101 ! + 1 102 ! + 2 103 1 101 ! + 1 102 ! + 1 103 ! + … + 2 200 1 101 ! + 1 102 ! + … + 1 200 ! 2 101 + … + 2 200 101 ! + 2 102 + … + 2 200 102 ! + … + 2 200 200 ! = 2 101 2 100 - 1 101 ! + 2 102 2 99 - 1 102 ! + … + 2 200 200 ! = 2 201 101 ! - 2 101 101 ! + 2 201 102 ! - 2 102 102 ! + … + 2 201 200 ! - 2 200 200 ! = ∑ n = 101 200 2 201 - 2 n n ! = b ∴ a b = 1

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