(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 $n \in {\mathbb{Z}^ + }$.

(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]

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.