A sequence has a_1 = 1, a_2 = 2 and a_n = 3a_(n-1) + 4a_(n-2) for n >= 3. What is its sixth term?

1. Two initial terms are given, so the rule can be applied from n = 3 onwards. 2. a_3 = 3(2) + 4(1) = 6 + 4 = 10. 3. a_4 = 3(10) + 4(2) = 30 + 8 = 38. 4. a_5 = 3(38) + 4(10) = 114 + 40 = 154. 5. a_6 = 3(154) + 4(38) = 462 + 152 = 614. 6. The value 154 is the fifth term, so stopping one step early is the commonest slip here. 7. Using 3(154) + 4(10) instead would give 502 and using only 3(154) + 38 would give 500, so the two preceding terms must both be the immediate ones. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 13 "Sequences, Probability and Counting Theory", section 13.1 Sequences and Their Notations_