Question

Using mathematical induction, the numbers \( a_n \)'s 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. \( 3n^3 + 3n + 1 \)
• C. \( n^3 - n^2 + 1 \)
• D. \( n^3 + n^2 \)

Hint:

When solving recursive sequences using induction, always start by proving the base case, then assume the formula for \( n = k \), and finally prove it for \( n = k+1 \).

Answer 1

The Correct Option is B

We are given that the sequence is defined recursively as: \[ a_0 = 1, \quad a_{n+1} = 3n^2 + n + a_n \] We need to find a general expression for \( a_n \). Step 1: Check the base case For \( n = 0 \): \[ a_0 = 1 \quad \text{(which is true)} \] Step 2: Assume the formula holds for \( n = k \) Assume that \( a_k = 3k^3 + 3k + 1 \) holds for some \( k \). Step 3: Prove for \( n = k+1 \) Using the recursive formula: \[ a_{k+1} = 3k^2 + k + a_k \] Substitute the assumed formula for \( a_k \): \[ a_{k+1} = 3k^2 + k + (3k^3 + 3k + 1) \] Simplify: \[ a_{k+1} = 3k^3 + 3k^2 + 3k + k + 1 = 3k^3 + 3k^2 + 3k + 1 \] Thus, the formula holds, and the general expression for \( a_n \) is: \[ a_n = 3n^3 + 3n + 1 \] Thus, the correct answer is \( 3n^3 + 3n + 1 \).