Question

If a₁ = 1, a₂ = 1 and a_n = 2a_{n - 1} + a_{n - 2}, n ≥ 3, n ∈ N, then find the first six terms of the sequence.

Answer 1

a₁ = 1, a₂ = 1, a_n = 2a_{n - 1} + a_{n - 2} 

a₃ = 2a_{(3 - 1)} + a_{(3 - 2)} 

= 2a₂ + a₁

= 2 x 1 + 1 = 3 

a₄ = 2a_{(4 - 1)} + a_{(4 - 2)} 

= 2a₃ + a₂ 

= 2 x 3 + 1 = 7 

a₅ = 2a_{(5 - 1)} + a_{(5 - 2)} 

= 2a₄ + a₃ 

= 2 x 7 + 3 = 17 

a₆ = 2a_{(6 - 1)} + a_{(6 - 2)} 

= 2a₅ + a₄ 

= 2 x 17 + 7 

= 34 + 7 

= 41

∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 …