A-LevelFurther MathematicsProof by induction9231/12 · May/June 2011 · Q46 Marks

It is given that \( f(n) = 3^{3n} + 6^{n-1} \).

(i)Show that \( f(n + 1) + f(n) = 28(3^{3n}) + 7(6^{n-1}) \).[2]
(ii)Hence, or otherwise, prove by mathematical induction that \( f(n) \) is divisible by 7 for every positive integer n.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.

[M1A1] Establishes initial result. \( f(n)+f(n+1) = 3^{3n} +6^{n-1} +3^{3n+3} +6^n \) \( = 3^{3n} (1 + 27) + 6^{n-1} (1 + 6) \) \( = 28(3^{3n}) + 7(6^{n-1}) \) (AG)
[B1] States inductive hypothesis. \( H_k: f(k) = 7\lambda \)
[B1] Proves base case. \( 3^3 + 6^0 = 28 = 4\times7 \Rightarrow H_1 \) is true
[M1] Shows P_(k) ⇒ P_(k+1) . f(k + 1) + f(k) = f(k+1) + 7λ = 28(3^(3k)) +7(6^(k-1)) = 7μ ⇒ f(k + 1) = 7(μ – λ) ... H_(k) ⇒ H_(k+1)
[A1] States conclusion. (Hence by the principle of mathematical induction \( H_n \) is) true for all positive integers n.

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Topic: Proof by induction, A-Level Further Mathematics (syllabus 9231). Structured question, 6 marks.
Source: A-Level Further Mathematics, May/June 2011 Paper 1 Variant 2, question 4.

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