A sequence {Vₙ} is defined for n ≥ 1 by: I)

Solution

Let V₁ = p, V₂ = q. From recurrence for n = 3: V₃ = 2V₂ − V₁ + 4 22 = 2q − p + 4 2q − p = 18 …(1) Find general expressions: V₄ = 2V₃ − V₂ + 4 = 2(2q − p + 4) − q + 4 = 4q − 2p + 8 − q + 4 = 3q − 2p + 12 V₅ = 2V₄ − V₃ + 4 = 2(3q − 2p + 12) − (2q − p + 4) + 4 = (6q − 4p + 24) − 2q + p − 4 + 4 = 4q − 3p + 24 Given V₅ = 34: 4q − 3p + 24 = 34 4q − 3p = 10 …(2) Now solve (1) and (2): From (1): 2q − p = 18 Multiply by 2: 4q − 2p = 36 …(3) Subtract (2) from (3): (4q − 2p) − (4q − 3p) = 36 − 10 p = 26 Substitute in (1): 2q − 26 = 18 2q = 44 q = 22 Now find V₄ and V₆. V₄ = 3q − 2p + 12 = 3×22 − 2×26 + 12 = 66 − 52 + 12 = 26 V₅ (check) = 34 (consistent) V₆ = 2V₅ − V₄ + 4 = 2×34 − 26 + 4 = 68 − 26 + 4 = 46 So V₆ = 46.