Question

the sequence given is defined using a recursion formula. write the first four terms of the se
a₁ = 4 and aₙ = 3aₙ₋₁ for n ≥ 2

a₁ = 4 (simplify your answer.)
a₂ = □ (simplify your answer.)

Explanation:

Step1: Find \(a_2\)

Given \(a_n = 3a_{n - 1}\) for \(n\geq2\) and \(a_1 = 4\). For \(n = 2\), substitute \(n = 2\) into the recursion formula: \(a_2=3a_{2 - 1}=3a_1\). Since \(a_1 = 4\), then \(a_2 = 3\times4 = 12\).

Step2: Find \(a_3\)

For \(n = 3\), use the recursion formula \(a_3 = 3a_{3 - 1}=3a_2\). We know \(a_2 = 12\), so \(a_3 = 3\times12 = 36\).

Step3: Find \(a_4\)

For \(n = 4\), use the recursion formula \(a_4 = 3a_{4 - 1}=3a_3\). We know \(a_3 = 36\), so \(a_4 = 3\times36 = 108\).

Answer:

The first four terms are \(a_1 = 4\), \(a_2 = 12\), \(a_3 = 36\), \(a_4 = 108\). (For the \(a_2\) box in the problem, the answer is 12)