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COMEDK · Maths · 6. Mathematical Induction

Using mathematical induction, the numbers \(a_{n}\) 's are defined by \(a_{0}=1, a_{n+1}=3 n^{2}+n+a_{n}\), \((n \geq 0)\). Then, \(a_{n}\) is equal to

  1. A \(n^{3}+n^{2}+1\)
  2. B \(n^{3}-n^{2}+1\)
  3. C \(n^{3}-n^{2}\)
  4. D \(n^{3}+n^{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(n^{3}-n^{2}+1\)

Step-by-step Solution

Detailed explanation

Given the recurrence relation \(a_{n+1} = a_{n} + 3n^{2} + n\) with \(a_{0} = 1\). We can write the terms as: \(a_{n} = a_{0} + \sum_{k=0}^{n-1} (a_{k+1} - a_{k})\) \(a_{n} = 1 + \sum_{k=0}^{n-1} (3k^{2} + k)\) Using the standard summation formulas…