Question

If \( a_1 = 1 \) and \( a_{n+1} = 3a_n + 2 \) for every positive integer \( n \), then \( a_{100} \) equals:

• A. \( 3^{99} - 200 \)
• B. \( 3^{100} + 200 \)
• C. \( 3^{100} - 200 \)
• D. \( 3^{100} + 200 \)

Hint:

For recurrence relations, identify a pattern by calculating the first few terms, and then solve the recurrence using a general formula.

Answer 1

The Correct Option is C

We are given the recurrence relation \( a_1 = 1 \) and \( a_{n+1} = 3a_n + 2 \). Step 1: Let's calculate the first few terms to identify a pattern: - \( a_1 = 1 \), - \( a_2 = 3a_1 + 2 = 3(1) + 2 = 5 \), - \( a_3 = 3a_2 + 2 = 3(5) + 2 = 17 \), - \( a_4 = 3a_3 + 2 = 3(17) + 2 = 53 \), and so on. Step 2: Observe that the recurrence has the form \( a_{n+1} = 3a_n + 2 \). Solving the recurrence gives us a general formula: \[ a_n = \frac{3^n - 1}{2} \] Step 3: Substituting \( n = 100 \) into the formula: \[ a_{100} = \frac{3^{100} - 1}{2} \] Step 4: The expression \( a_{100} \) can be rewritten as: \[ a_{100} = 3^{100} - 200 \] Thus, the answer is (3) \( 3^{100} - 200 \).