A-LevelFurther MathematicsProof by induction9231/12 · May/June 2016 · Q36 Marks
Prove by mathematical induction that, for all positive integers n, 10ⁿ + 3 × 4ⁿ⁺² + 5 is divisible by 9.
- Prove by mathematical induction that, for all positive integers n, 10ⁿ + 3 × 4ⁿ⁺² + 5 is divisible by 9.[6]
Mark scheme
Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.
- [B1] For n=1 10+192 + 5 = 207=9×23 ⇒ H₁ is true.
- [B1] Assume Hk is true for some positive integer k ⇒10ᵏ +3.4ᵏ⁺² + 5 = 9α
- [M1] Hence f(k+1)-f(k)=10ᵏ (10-1)+3.4ᵏ⁺² (4-1)
- [A1] = 9(10ᵏ +4ᵏ⁺²)
- [A1] Hence f(k+1)(=9(β+α))⇒ Hᵏ⁺¹ is true
- [A1] H₁ is true and Hk ⇒ Hk+1, hence by PMI Hₙ is true for all positive integers n.
- N.B. Or can show f(n+1) = 9(10α -2.4ⁿ⁺²+5) for M1A1A1. (3rd, 4th&5th marks)
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