Answer only one of the following two alternatives. EITHER The roots of the equation x⁴ – 3x² + 5x − 2 = 0 are α, β, γ, δ, and αⁿ + βⁿ + γⁿ + δⁿ is denoted by S_n.
- Show that S_(n+4) – 3S_(n+2) + 5S_(n+1) – 2S_n = 0.[2]
- (i)S₂ and S₄,[3]
- (ii)S₃ and S₅.[6]
- Hence find the value of α²(β³ + γ³ + δ³) + β²(γ³ + δ³ + α³) + γ²(δ³ + α³ + β³) + δ²(α³ + β³ + γ³).[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2012). The mark or mark type is shown in brackets.
- [M1] Substitute α into equation. Multiply by αⁿ. Obtain result: α⁴ – 3α² + 5α – 2 = 0 ⇒ αⁿ⁺⁴ – 3αⁿ⁺² + 5αⁿ⁺¹ – 2αⁿ = 0. Repeat for β,γ,δ and sum.
- [A1] S_(n+4) – 3S_(n+2) + 5S_(n+1) – 2S_n = 0. (AG)
- [B1] Uses Σα² = (Σα)² - 2Σαβ. S₀ = 4. S₁ = Σα = 0. S₂ = (S₁)² - 2Σαβ = 0 - 2(-3) = 6.
- [M1A1] Finds S₄ from formula: S₄ = 3S₂ - 5S₁ + 2S₀ = 3(6) - 5(0) + 2(4) = 18 + 8 = 26.
- [M1A1] S_(-1) = (Σαβγ) / (αβγδ) = (-5) / (-2) = 5/2.
- [M1A1] Finds S₃ from formula: S₃ = 3S₁ - 5S₀ + 2S_(-1) = 3(0) - 5(4) + 2(5/2) = -20 + 5 = -15.
- [M1A1] Finds S₅ from formula: S₅ = 3S₃ - 5S₂ + 2S₁ = 3(-15) - 5(6) + 2(0) = -45 - 30 = -75.
- [M1] Σα²β³ = S₂S₃ - S₅
- [M1A1] = 6(-15) - (-75) = -90 + 75 = -15.
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