COMEDK · Maths · 6. Mathematical Induction
Using mathematical induction, the numbers \(a_n \delta\) 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 \(n^3-n^2\)
- B \(n^3+n^2+1\)
- C \(n^3-n^2+1\)
- D \(n^3+n^2\)
Answer & Solution
Correct Answer
(C) \(n^3-n^2+1\)
Step-by-step Solution
Detailed explanation
The recurrence relation is given by \(a_{n+1} = a_n + 3n^2 + n\) with \(a_0 = 1\). We can write the general term \(a_n\) as a summation: \(a_n = a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) = 1 + \sum_{k=0}^{n-1} (3k^2 + k)\). Using the standard summation formulas…
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