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Date May 2010 Marks available 10 Reference code 10M.1.hl.TZ2.11
Level HL only Paper 1 Time zone TZ2
Command term Prove Question number 11 Adapted from N/A

Question

(a)     Consider the following sequence of equations.

     \(1 \times 2 = \frac{1}{3}(1 \times 2 \times 3),\)

     \(1 \times 2 + 2 \times 3 = \frac{1}{3}(2 \times 3 \times 4),\)

     \(1 \times 2 + 2 \times 3 + 3 \times 4 = \frac{1}{3}(3 \times 4 \times 5),\)

     \( \ldots {\text{ .}}\)

(i)     Formulate a conjecture for the \({n^{{\text{th}}}}\) equation in the sequence.

(ii)     Verify your conjecture for n = 4 .

(b)     A sequence of numbers has the \({n^{{\text{th}}}}\) term given by \({u_n} = {2^n} + 3,{\text{ }}n \in {\mathbb{Z}^ + }\). Bill conjectures that all members of the sequence are prime numbers. Show that Bill’s conjecture is false.

(c)     Use mathematical induction to prove that \(5 \times {7^n} + 1\) is divisible by 6 for all \(n \in {\mathbb{Z}^ + }\).

Markscheme

(a)     (i)     \(1 \times 2 + 2 \times 3 + \ldots + n(n + 1) = \frac{1}{3}n(n + 1)(n + 2)\)     R1

 

(ii)     LHS = 40; RHS = 40     A1

[2 marks]

 

(b)     the sequence of values are:

5, 7, 11, 19, 35 … or an example     A1

35 is not prime, so Bill’s conjecture is false     R1AG

[2 marks]

 

(c)     \({\text{P}}(n):5 \times {7^n} + 1\) is divisible by 6

\({\text{P}}(1):36\) is divisible by \(6 \Rightarrow {\text{P}}(1)\) true     A1

assume \({\text{P}}(k)\) is true \((5 \times {7^k} + 1 = 6r)\)     M1

Note: Do not award M1 for statement starting ‘let n = k’.

Subsequent marks are independent of this M1.

 

consider \(5 \times {7^{k + 1}} + 1\)     M1

\( = 7(6r - 1) + 1\)     (A1)

\( = 6(7r - 1) \Rightarrow {\text{P}}(k + 1)\) is true     A1

P(1) true and \({\text{P}}(k)\) true \( \Rightarrow {\text{P}}(k + 1)\) true, so by MI \({\text{P}}(n)\) is true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note: Only award R1 if there is consideration of P(1), \({\text{P}}(k)\) and \({\text{P}}(k + 1)\) in the final statement.

Only award R1 if at least one of the two preceding A marks has been awarded.

 

[6 marks]

Total [10 marks]

Examiners report

Although there were a good number of wholly correct solutions to this question, it was clear that a number of students had not been prepared for questions on conjectures. The proof by induction was relatively well done, but candidates often showed a lack of rigour in the proof. It was fairly common to see students who did not appreciate the idea that \({\text{P}}(k)\) is assumed not given and this was penalised. Also it appeared that a number of students had been taught to write down the final reasoning for a proof by induction, even if no attempt of a proof had taken place. In these cases, the final reasoning mark was not awarded.

Syllabus sections

Topic 1 - Core: Algebra » 1.4 » Proof by mathematical induction.

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