AS June 2018 Paper 1 Q8
8.
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1,\ \begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}^1 = \begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix},\quad \begin{pmatrix}4 \times 1 + 1 & -8(1)\\ 2 \times 1 & 1 - 4(1)\end{pmatrix} = \begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}\) So the result is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(\begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}^k = \begin{pmatrix}4k + 1 & -8k\\ 2k & 1 - 4k\end{pmatrix}\) | M1 | 2.4 |
| \[\begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}^{k+1} = \begin{pmatrix}4k + 1 & -8k\\ 2k & 1 - 4k\end{pmatrix}\begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}\]or\[\begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}^{k+1} = \begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}\begin{pmatrix}4k + 1 & -8k\\ 2k & 1 - 4k\end{pmatrix}\] | M1 | 1.1b |
| \[\begin{pmatrix}4k + 1 & -8k\\ 2k & 1 - 4k\end{pmatrix}\begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix} = \begin{pmatrix}5(4k + 1) - 16k & -8(4k + 1) + 24k\\ 10k + 2(1 - 4k) & -16k - 3(1 - 4k)\end{pmatrix}\]or\[\begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}\begin{pmatrix}4k + 1 & -8k\\ 2k & 1 - 4k\end{pmatrix} = \begin{pmatrix}5(4k + 1) - 16k & -40k - 8(1 - 4k)\\ 2(1 + 4k) - 6k & -16k - 3(1 - 4k)\end{pmatrix}\] | A1 | 1.1b |
| \[= \begin{pmatrix}4(k + 1) + 1 & -8(k + 1)\\ 2(k + 1) & 1 - 4(k + 1)\end{pmatrix}\] | A1 | 2.1 |
| If true for \(n = k\) then true for \(n = k + 1\), true for \(n = 1\) so true for all (positive integers) \(n\) (Allow “for all values”) | A1 | 2.4 |
| (6) |
Notes
B1: Shows that the result holds for \(n = 1\). Must see substitution into the rhs.
The minimum would be: \(\begin{pmatrix}4 + 1 & -8\\ 2 & 1 - 4\end{pmatrix} = \begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}\).
M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)
M1: Sets up a correct multiplication statement either way round
A1: Achieves a correct un-simplified matrix
A1: Reaches a correct simplified matrix with no errors and the correct un-simplified matrix seen previously. Note that the simplified result may be proved by equivalence.
A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.
Way 1: \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(4^{n+1} + 5^{2n-1} = 16 + 5 = 21\) so the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(4^{k+1} + 5^{2k-1}\) is divisible by 21 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 4^{k+2} + 5^{2k+1} - 4^{k+1} - 5^{2k-1}\) | M1 | 2.1 |
| \(= 4 \times 4^{k+1} + 25 \times 5^{2k-1} - 4^{k+1} - 5^{2k-1}\) | ||
| \(= 3\mathrm{f}(k) + 21 \times 5^{2k-1}\) or e.g. \(= 24\mathrm{f}(k) - 21 \times 4^{k+1}\) | A1 | 1.1b |
| \(\mathrm{f}(k + 1) = 4\mathrm{f}(k) + 21 \times 5^{2k-1}\) or e.g. \(\mathrm{f}(k + 1) = 25\mathrm{f}(k) - 21 \times 4^{k+1}\) | A1 | 1.1b |
| If true for \(n = k\) then true for \(n = k + 1\), true for \(n = 1\) so true for all (positive integers) \(n\) (Allow “for all values”) | A1 | 2.4 |
| (6) | ||
| (12 marks) |
Notes
(ii) Way 1
B1: Shows that f(1) = 21
M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)
M1: Attempts f(\(k\) + 1) – f(\(k\)) or equivalent work
A1: Achieves a correct expression for f(\(k\) + 1) – f(\(k\)) in terms of f(\(k\))
A1: Reaches a correct expression for f(\(k\) + 1) in terms of f(\(k\))
A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.
Alternative: Way 2: \(\mathrm{f}(k + 1)\)
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(4^{n+1} + 5^{2n-1} = 16 + 5 = 21\) so the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(4^{k+1} + 5^{2k-1}\) is divisible by 21 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) = 4^{k+1+1} + 5^{2(k+1)-1}\) | M1 | 2.1 |
| \(\mathrm{f}(k + 1) = 4 \times 4^{k+1} + 5^{2k+1} = 4 \times 4^{k+1} + 4 \times 5^{2k-1} + 25 \times 5^{2k-1} - 4 \times 5^{2k-1}\) \(\mathrm{f}(k + 1) = 4\mathrm{f}(k) + 21 \times 5^{2k-1}\) | A1 A1 | 1.1b 1.1b |
| If true for \(n = k\) then true for \(n = k + 1\), true for \(n = 1\) so true for all (positive integers) \(n\) (Allow “for all values”) | A1 | 2.4 |
| (6) |
B1: Shows that f(1) = 21
M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)
M1: Attempts f(\(k\) + 1)
A1: Correctly obtains 4f(\(k\)) or \(21 \times 5^{2k-1}\)
A1: Reaches a correct expression for f(\(k\) + 1) in terms of f(\(k\))
A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.
Alternative: Way 3: \(\mathrm{f}(k + 1) - m\mathrm{f}(k)\)
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(4^{n+1} + 5^{2n-1} = 16 + 5 = 21\) so the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(4^{k+1} + 5^{2k-1}\) is divisible by 21 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) - m\mathrm{f}(k) = 4^{k+2} + 5^{2k+1} - m\left(4^{k+1} + 5^{2k-1}\right)\) | M1 | 2.1 |
| \(= (4 - m)4^{k+1} + 5^{2k+1} - m \times 5^{2k-1}\) \(= (4 - m)\left(4^{k+1} + 5^{2k-1}\right) + 21 \times 5^{2k-1}\) | A1 | 1.1b |
| \(= (4 - m)\left(4^{k+1} + 5^{2k-1}\right) + 21 \times 5^{2k-1} + m\mathrm{f}(k)\) | A1 | 1.1b |
| If true for \(n = k\) then true for \(n = k + 1\), true for \(n = 1\) so true for all (positive integers) \(n\) (Allow “for all values”) | A1 | 2.4 |
| (6) |
B1: Shows that f(1) = 21
M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)
M1: Attempts f(\(k\) + 1) – \(m\)f(\(k\))
A1: Achieves a correct expression for f(\(k\) + 1) – \(m\)f(\(k\)) in terms of f(\(k\))
A1: Reaches a correct expression for f(\(k\) + 1) in terms of f(\(k\))
A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.
Alternative: Way 4: \(\mathrm{f}(k) = 21M\)
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(4^{n+1} + 5^{2n-1} = 16 + 5 = 21\) so the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(4^{k+1} + 5^{2k-1} = 21M\) | M1 | 2.4 |
| \(\mathrm{f}(k + 1) = 4^{k+1+1} + 5^{2(k+1)-1}\) | M1 | 2.1 |
| \(\mathrm{f}(k + 1) = 4 \times 4^{k+1} + 5^{2k+1} = 4\left(21M - 5^{2k-1}\right) + 5^{2k+1}\) \(\mathrm{f}(k + 1) = 84M + 21 \times 5^{2k-1}\) | A1 A1 | 1.1b 1.1b |
| If true for \(n = k\) then true for \(n = k + 1\), true for \(n = 1\) so true for all (positive integers) \(n\) (Allow “for all values”) | A1 | 2.4 |
| (6) |
B1: Shows that f(1) = 21
M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)
M1: Attempts f(\(k\) + 1)
A1: Correctly obtains \(84M\) or \(21 \times 5^{2k-1}\)
A1: Reaches a correct expression for f(\(k\) + 1) in terms of \(M\) and \(5^{2k-1}\)
A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.