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NTA Abhyas JEE Main2020MathematicsBinomial TheoremPractice

∑ r = 0 n r 2 r + 1 n C r is equal to

Options

  1. A2 n - 1 n 2 + n + 2 - 1 n + 1
  2. B2 n - 1 n 2 - n - 2 + 1 n + 1
  3. C2 n - 1 n 2 - n + 2 - 1 n + 1
  4. D2 n - 1 n 2 + n - 2 + 1 n + 1

Correct answer

C. 2 n - 1 n 2 - n + 2 - 1 n + 1

Step-by-step solution

∑ r = 0 n r 2 r + 1 n C r = ∑ r = 0 n r 2 - 1 + 1 r + 1 n C r = ∑ r = 0 n r - 1 + 1 r + 1 n C r = ∑ r = 0 n r ⋅ n C r - ∑ r = 0 n n C r + ∑ r = 0 n 1 r + 1 n C r 1 r + 1 n C r = 1 n + 1 n + 1 C r + 1   and   r ⋅ n C r = n · n - 1 C r - 1 = ∑ r = 1 n n · n - 1 C r - 1 - 2 n + ∑ r = 0 n 1 n + 1 · n + 1 C r + 1 = n n - 1 C 0 + n - 1 C 1 + . . . + n - 1 C n - 1 - 2 n + 1 n + 1 n + 1 C 1 + n + 1 C 2 + . . . + n + 1 C n + 1 = n ⋅ 2

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