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.