June 2019 Paper 2 Q8
8
(a) A linear sequence has first term \(\quad 7 + 12\sqrt{5}\)
The term-to-term rule is
add \(\ 9 - 2\sqrt{5}\)
One term of the sequence is an integer.
Work out the value of this integer. [2 marks]
(b) The \(n\)th term of a different sequence is \(\quad \dfrac{3n^2 - 1}{n^2 + 1}\)
Work out the sum of the first three terms. [2 marks]
(c) The first four terms of a quadratic sequence are
\(-3 \qquad\quad 3 \qquad\quad 13 \qquad\quad 27\)
Work out an expression for the \(n\)th term. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(7 + 12\sqrt{5} + 6(9 - 2\sqrt{5})\) or \(12\sqrt{5} + 6(-2\sqrt{5}) = 0\) or \(12\sqrt{5} \div 2\sqrt{5} = 6\) or states that need to add 6 lots of \((9 - 2\sqrt{5})\) or 7th term | M1 | oe eg \(7 + 6 \times 9\) or \(7 + 54\) or \(6 \times -2 = -12\) allow \(7 + 12\sqrt{5} + (n - 1)(9 - 2\sqrt{5})\) with \(n = 7\) allow \(7 + 12\sqrt{5} + n(9 - 2\sqrt{5})\) with \(n = 6\) |
| 61 | A1 |
Additional guidance
| 61 in working lines with 7(th) on answer line | M1A0 |
| If repeatedly adding \((9 - 2\sqrt{5})\) they must stop after adding 6 lots or clearly select the relevant one | |
| Answer 6 or 6th term with M1 not seen | M0A0 |
| Ignore any conversions to decimals | |
| Beware \(\ (9 - 2\sqrt{5})(9 + 2\sqrt{5}) = 61\) | M0A0 |
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{29}{5}\) or \(5\dfrac{4}{5}\) or 5.8 | B2 | oe eg \(5\dfrac{8}{10}\) B1 any two of 1, \(\dfrac{11}{5}\), \(\dfrac{26}{10}\) oe values |
Additional guidance
| Terms must be evaluated for B1 unless correct answer seen eg1 \(\ \dfrac{3 - 1}{1 + 1} + \dfrac{12 - 1}{4 + 1} + \dfrac{27 - 1}{9 + 1}\) eg2 \(\ \dfrac{3 - 1}{1 + 1} + \dfrac{12 - 1}{4 + 1} + \dfrac{27 - 1}{9 + 1} = 5.8\) | B0 B2 |
| 1 \(\quad\) 7 \(\quad\) 2.6 | B1 |
| Ignore conversion attempts after a correct value seen |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| (Second differences \(=\)) 4 or \(2n^2\) | M1 | second differences seen at least once and not contradicted may be seen by the sequence |
| \(-3 - 2 \quad 3 - 8 \quad (13 - 18 \quad 27 - 32)\) or \(-5 \quad -5 \quad (-5 \quad -5)\) | M1dep | subtracts \(2n^2\) from the given terms |
| \(2n^2 - 5\) | A1 | oe eg \(2n^2 + 0n - 5\) does not need terms collected |
| Alternative method 2 | ||
| (Second differences \(=\)) 4 or \(2n^2\) | M1 | second differences seen at least once and not contradicted may be seen by the sequence |
| \(3a + b = 3 - -3\) and substitutes \(a = 2\) or \(b = 0\) | M1dep | oe |
| \(2n^2 - 5\) | A1 | oe eg \(2n^2 + 0n - 5\) does not need terms collected |
| Alternative method 3 | ||
| Any three of \(a + b + c = -3\) \(4a + 2b + c = 3\) \(9a + 3b + c = 13\) \(16a + 4b + c = 27\) | M1 | |
| \(3a + b = 3 - -3\) and \(5a + b = 13 - 3\) or \(a = 2\) and \(b = 0\) | M1dep | oe obtains two correct equations in same two variables from their equations |
| \(2n^2 - 5\) | A1 | oe eg \(2n^2 + 0n - 5\) does not need terms collected |
| Alternative method 4 | ||
| (Second differences \(=\)) 4 or \(2n^2\) | M1 | second differences seen at least once and not contradicted may be seen by the sequence |
| \(2 \times 1^2 + b \times 1 - 5 = -3\) or \(2 + b - 5 = -3\) or \(b = 0\) | M1dep | \(2n^2 + bn - 5 = -3\) with \(n = 1\) substituted oe eg \(2 \times 2^2 + b \times 2 - 5 = 3\) |
| \(2n^2 - 5\) | A1 | oe eg \(2n^2 + 0n - 5\) does not need terms collected |
Additional guidance
| Condone working in a different variable | |
| Alt 1 2nd M1 Subtracting given terms from \(2n^2\) leading to 5 5 (5 5) must be recovered eg final answer \(2n^2 - 5\) (\(2n^2 - 5n\) or \(2n^2 - 5n - 5\) is not a recovery) | |
| Answer \(2n^2\) …… scores at least M1 | |
| Condone \(\ n = 2n^2 - 5\) or \(2n^2 - 5 = 0\) | M2A1 |