If a1 = 1, a2 = 1 and an = 2an - 1 + an - 2, n ≥ 3, n ∈ N, then f

1.	If a1 = 1, a2 = 1 and an = 2an - 1 + an - 2, n ≥ 3, n ∈ N, then find the first six terms of the sequence.
Answer» 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 …

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

Answer»

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 …