Higher November 2022 Paper 4 Q18
18
(a) The next term in a Fibonacci sequence is found by adding together the two previous terms.
(i) The first and second terms of a particular Fibonacci sequence are \(x\) and \(y\).
Show that the fourth term of the sequence can be written as \(x + 2y\). [2]
Show that the fourth term of the sequence can be written as \(x + 2y\). [2]
(ii) The fourth term of the same Fibonacci sequence is 7.
The seventh term of the sequence is 31.
Work out the value of \(x\) and the value of \(y\).
You must show your working. [6]
The seventh term of the sequence is 31.
Work out the value of \(x\) and the value of \(y\).
You must show your working. [6]
(b) Here are the first four terms of a sequence.\[1 \qquad \sqrt{3} \qquad 3 \qquad 3\sqrt{3}\]
Write an expression for the \(n\)th term. [2]
(c) Here are the first four terms of a quadratic sequence.\[-1 \qquad 5 \qquad 13 \qquad 23\]
The \(n\)th term is \(n^2 + bn + c\).
Find the value of \(b\) and the value of \(c\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (a)(i) | |||
| [\(u_3\) = ] \(x + y\) | 1 | ||
| [\(u_4\) = ] \(y + x + y\) [= \(x + 2y\)] | 1 | ||
| (a)(ii) | |||
| 3 2 with correct working | 6 | M2 for [\(u_7\) = ] \(7 + x + y + 7 + x + y + 7\) oe or better or M1 for [\(u_5\) = ] \(x + y + 7\) oe or better and B1 for \(x + 2y = 7\) oe B1FT for their \((2x + 2y + 21) = 31\) oe M1 for solving their equations e.g. subtracting equations to give \(x = 3\) If 0 or 1 scored, instead award SC2 for answers [\(x\) =] 3 and [\(y\) =] 2 with no or insufficient working If 0 scored, instead award SC1 for [\(x\) =] 3 or [\(y\) =] 2 or both correct answers switched, with no or insufficient working | “Correct working” requires evidence of at least B1 B1 or M2 or M1 M1 e.g. \(2x + 2y + 21\) or \(2x + 3y + x + 2y + 2x + 3y\) or \(5x + 8y\) e.g. \(x + y + x + 2y\) or \(2x + 3y\) e.g. \(5x + 8y = 31\) FT their \(u_7\) e.g. multiplying one equation and correctly adding or subtracting to eliminate one variable correct answers with trials will score 6 marks M1 for each correct trial (value of \(x\) and a value of \(y\)) up to a maximum of M3 e.g. \(x = 1\) \(y = 4\) gives 1 4 5 9 [14 23 37] |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((\sqrt{3})^{n-1}\) or \(3^{\frac{1}{2}(n-1)}\) oe | 2 | M1 for common ratio of \(\sqrt{3}\) implied by answer of \((\sqrt{3})^k\) | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3 −5 | 3 | B2 for \(b\) = 3 OR M1 for [−1 5 13 23] − [1 4 9 16] implied by −2 1 4 7 B1 for \(c\) = −5 | condone 3\(n\) for 2 marks If equations used M1 for e.g. 1 + b + c = −1 oe 4 + 2b + c = 5 oe Allow any correct method |