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∑ r = 1 n r r 4 + r 2 + 1 is equal to

Options

  1. An 2 + n 2 n 2 + n + 1
  2. Bn 2 + 2 n 2 n 2 + n + 1
  3. C2 n 2 + n 2 n 2 + n + 1
  4. Dn 2 + n n 2 + n + 1

Correct answer

A. n 2 + n 2 n 2 + n + 1

Step-by-step solution

Let, Ι r = r r 4 + r 2 + 1 = r r 2 + r + 1 r 2 - r + 1 = 1 2 r 2 + r + 1 - r 2 - r + 1 r 2 + r + 1 r 2 - r + 1 = - 1 2 1 r 2 + r + 1 - 1 r 2 - r + 1 = - 1 2 V r - V r - 1 ⇒ ∑ r = 1 n Ι r = - 1 2 V n - V 0 = - 1 2 1 n 2 + n + 1 - 1 = 1 2 1 - 1 n 2 + n + 1 = 1 2 n 2 + n n 2 + n + 1

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