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
- A \(n^{3}+n^{2}+1\)
- B \(n^{3}-n^{2}+1\)
- C \(n^{3}-n^{2}\)
- D \(n^{3}+n^{2}\)
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…
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