Question

Using mathematical induction, the numbers $ a_n $ are defined by $ a_0 = 1, a_{n+1} = 3n^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. \( n^3 + n^2 \)

Hint:

When using mathematical induction, always ensure you correctly handle the base case, make an assumption for \( k \), and prove the result for \( k + 1 \) using the recursive relation.

Answer 1

The Correct Option is B

We are given that \( a_0 = 1 \) and the recursive relation \( a_{n+1} = 3n^2 + n + a_n \). 
We need to find a formula for \( a_n \) using mathematical induction. 
Step 1: Base Case (Induction Start)
For \( n = 0 \), we have: \[ a_0 = 1 \] This satisfies the base case, so the induction hypothesis holds for \( n = 0 \). 
Step 2: Induction Hypothesis
Assume that the formula holds for some \( k \), i.e., assume that: \[ a_k = k^3 - k^2 + 1 \] 
Step 3: Inductive Step
We now need to prove that the formula holds for \( k + 1 \), i.e., we need to prove: \[ a_{k+1} = (k+1)^3 - (k+1)^2 + 1 \] Using the recursive formula: \[ a_{k+1} = 3k^2 + k + a_k \] Substitute the induction hypothesis for \( a_k \): \[ a_{k+1} = 3k^2 + k + (k^3 - k^2 + 1) \] Simplify the expression: \[ a_{k+1} = 3k^2 + k + k^3 - k^2 + 1 \] \[ a_{k+1} = k^3 + 2k^2 + k + 1 \] Now, expand \( (k+1)^3 - (k+1)^2 + 1 \): \[ (k+1)^3 - (k+1)^2 + 1 = (k^3 + 3k^2 + 3k + 1) - (k^2 + 2k + 1) + 1 \] \[ = k^3 + 3k^2 + 3k + 1 - k^2 - 2k - 1 + 1 \] \[ = k^3 + 2k^2 + k + 1 \] Thus, \( a_{k+1} = (k+1)^3 - (k+1)^2 + 1 \), which completes the inductive step. 
Therefore, by the principle of mathematical induction, the formula holds for all \( n \geq 0 \), and we have: \[ a_n = n^3 - n^2 + 1 \]