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Question
If a1 = 1, a2 = 1 and an = 2an−1 + an−2, n ≥ 3, n ∈ N, then find the first six terms of the sequence
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Solution
a1 = 1, a2 = 1, an = 2an−1 + an−2
a3 = 2a(3−1) + a(3−2)
= 2a2 + a1
= 2 × 1 + 1 = 3
a4 = 2a(4−1) + a(4−2)
= 2a3 + a2
= 2 × 3 + 1 = 7
a5 = 2a(5−1) + a(5−2)
= 2a4 + a3
= 2 × 7 + 3 = 17
a6 = 2a(6−1) + a(6−2)
= 2a5 + a4
= 2 × 17 + 7
= 34 + 7
= 41
∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 .......
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