A2 June 2019 Paper 1 Q6
6. Prove by induction that for all positive integers \(n\)
\[\mathrm{f}(n) = 3^{2n+4} - 2^{2n}\]is divisible by 5 (6)
Way 1 \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(3^{2n+4} - 2^{2n} = 729 - 4 = 725\) \((725 = 145 \times 5)\) so the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(3^{2k+4} - 2^{2k}\) is divisible by 5 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 3^{2k+6} - 2^{2k+2} - 3^{2k+4} + 2^{2k}\) | M1 | 2.1 |
| either \(8\mathrm{f}(k) + 5 \times 2^{2k}\) or \(3\mathrm{f}(k) + 5 \times 3^{2k+4}\) | A1 | 1.1b |
| \(\mathrm{f}(k + 1) = 9\mathrm{f}(k) + 5 \times 2^{2k}\) or \(\mathrm{f}(k + 1) = 4\mathrm{f}(k) + 5 \times 3^{2k+4}\) o.e. | A1 | 1.1b |
| If true for \(n = k\) then it is true for \(n = k + 1\) and as it is true for \(n = 1\), the statement is true for all (positive integers) \(n\). (Allow ‘for all values’) | A1 | 2.4 |
| (6) | ||
| (6 marks) |
Notes
Way 1 \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)
B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)
M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc
M1: Attempts \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) or equivalent work
A1: Achieves a correct simplified expression for \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)
A1: Achieves a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)
A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all 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 \(n = 1\), \(3^{2n+4} - 2^{2n} = 729 - 4 = 725\) \((725 = 145 \times 5)\) so the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(3^{2k+4} - 2^{2k}\) is divisible by 5 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) = 3^{2(k+1)+4} - 2^{2(k+1)}\ \left(= 3^{2k+6} - 2^{2k+2}\right)\) | M1 | 2.1 |
| \(\mathrm{f}(k + 1) = 9\mathrm{f}(k) + 5 \times 2^{2k}\) or \(\mathrm{f}(k + 1) = 4\mathrm{f}(k) + 5 \times 3^{2k+4}\) o.e. | A1 A1 | 1.1b 1.1b |
| If true for \(n = k\) then it is true for \(n = k + 1\) and as it is true for \(n = 1\), the statement is true for all (positive integers) \(n\). (Allow ‘for all values’) | A1 | 2.4 |
| (6) |
B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)
M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc
M1: Attempts \(\mathrm{f}(k + 1)\)
A1: Correctly achieves either \(9\mathrm{f}(k)\) or \(5 \times 2^{2k}\) or either \(4\mathrm{f}(k)\) or \(5 \times 3^{2k+4}\)
A1: Achieves a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)
A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.
Alternative: Way 3 \(\mathrm{f}(k) = 5M\)
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(3^{2n+4} - 2^{2n} = 729 - 4 = 725\) \((725 = 145 \times 5)\) so the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(3^{2k+4} - 2^{2k} = 5M\) | M1 | 2.4 |
| \(\mathrm{f}(k + 1) = 3^{2(k+1)+4} - 2^{2(k+1)}\ \left(= 3^{2k+6} - 2^{2k+2}\right)\) | M1 | 2.1 |
| \(\left(\mathrm{f}(k + 1) = 3^2 \times 3^{2k+4} - 2^2 \times 2^{2k} = 3^2 \times \left(5M + 2^{2k}\right) - 2^2 \times 2^{2k}\right)\) \(\mathrm{f}(k + 1) = 45M + 5 \times 2^{2k}\) o.e. OR \(\left(\mathrm{f}(k + 1) = 3^2 \times 3^{2k+4} - 2^2 \times 2^{2k} = 3^2 \times 3^{2k+4} - 2^2 \times \left(3^{2k+4} - 5M\right)\right)\) \(\mathrm{f}(k + 1) = 5 \times 3^{2k+4} + 20M\) o.e. | A1 A1 | 1.1b 1.1b |
| If true for \(n = k\) then it is true for \(n = k + 1\) and as it is true for \(n = 1\), the statement is true for all (positive integers) \(n\). (Allow ‘for all values’) | A1 | 2.4 |
| (6) |
(Corrected from the printed mark scheme: the first line is printed as \(3^2 \times \left(5M + 2^{2k+2}\right)\); since \(3^{2k+4} = 5M + 2^{2k}\), it should be \(3^2 \times \left(5M + 2^{2k}\right)\).)
B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)
M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc
M1: Attempts \(\mathrm{f}(k + 1)\)
A1: Correctly achieves either \(45M\) or \(5 \times 2^{2k}\) or either \(20M\) or \(5 \times 3^{2k+4}\)
A1: Achieves a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(M\) and \(2^{2k}\) or \(M\) and \(3^{2k+4}\)
A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.
Alternative: Way 4 \(\mathrm{f}(k + 1) + \mathrm{f}(k)\)
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(3^{2n+4} - 2^{2n} = 729 - 4 = 725\) \((725 = 145 \times 5)\) so the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(3^{2k+4} - 2^{2k}\) is divisible by 5 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) + \mathrm{f}(k) = 3^{2k+6} - 2^{2k+2} + 3^{2k+4} - 2^{2k}\) | M1 | 2.1 |
| \(\mathrm{f}(k + 1) + \mathrm{f}(k) = 3^2 \times 3^{2k+4} - 2^2 \times 2^{2k} + 3^{2k+4} - 2^{2k}\) leading to \(10 \times 3^{2k+4} - 5 \times 2^{2k}\) | A1 | 1.1b |
| \(\mathrm{f}(k + 1) = 5\left[2 \times 3^{2k+4} - 2^{2k}\right] - \mathrm{f}(k)\) o.e. | A1 | 1.1b |
| If true for \(n = k\) then it is true for \(n = k + 1\) and as it is true for \(n = 1\), the statement is true for all (positive integers) \(n\). (Allow ‘for all values’) | A1 | 2.4 |
| (6) |
B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)
M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc
M1: Attempts \(\mathrm{f}(k + 1) + \mathrm{f}(k)\) or equivalent work
A1: Achieves a correct simplified expression for \(\mathrm{f}(k + 1) + \mathrm{f}(k)\)
A1: Achieves a correct expression for \(\mathrm{f}(k + 1) = 5\left[2 \times 3^{2k+4} - 2^{2k}\right] - \mathrm{f}(k)\)
A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.
Alternative: Way 5 \(\mathrm{f}(k + 1) - \text{‘}M\text{’}\mathrm{f}(k)\)
(Selecting a value of M that will lead to multiples of 5)
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(3^{2n+4} - 2^{2n} = 729 - 4 = 725\) \((725 = 145 \times 5)\) so the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(3^{2k+4} - 2^{2k}\) is divisible by 5 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) - \text{‘}M\text{’}\mathrm{f}(k) = 3^{2k+6} - 2^{2k+2} - \text{‘}M\text{’} \times 3^{2k+4} + \text{‘}M\text{’} \times 2^{2k}\) | M1 | 2.1 |
| \(\mathrm{f}(k + 1) - \text{‘}M\text{’}\mathrm{f}(k) = \left(9 - \text{‘}M\text{’}\right) \times 3^{2k+4} - \left(4 - \text{‘}M\text{’}\right) \times 2^{2k}\) | A1 | 1.1b |
| \(\mathrm{f}(k + 1) = \left(9 - \text{‘}M\text{’}\right) \times 3^{2k+4} - \left(4 - \text{‘}M\text{’}\right) \times 2^{2k} + \text{‘}M\text{’}\mathrm{f}(k)\) o.e. | A1 | 1.1b |
| If true for \(n = k\) then it is true for \(n = k + 1\) and as it is true for \(n = 1\), the statement is true for all (positive integers) \(n\). (Allow ‘for all values’) | A1 | 2.4 |
| (6) |
Way 5 \(\mathrm{f}(k + 1) - M\mathrm{f}(k)\) (Selects a suitable value for M which leads to divisibility of 5)
B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)
M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc
M1: Attempts \(\mathrm{f}(k + 1) - M\mathrm{f}(k)\) or equivalent work
A1: Achieves a correct simplified expression, \(\mathrm{f}(k + 1) - \text{‘}M\text{’}\mathrm{f}(k)\) which is divisible by 5
\(\mathrm{f}(k + 1) - \text{‘}M\text{’}\mathrm{f}(k) = \left(9 - \text{‘}M\text{’}\right) \times 3^{2k+4} - \left(4 - \text{‘}M\text{’}\right) \times 2^{2k}\)
A1: Achieves a correct expression for \(\mathrm{f}(k + 1) = \left(9 - \text{‘}M\text{’}\right) \times 3^{2k+4} - \left(4 - \text{‘}M\text{’}\right) \times 2^{2k} + \text{‘}M\text{’}\mathrm{f}(k)\) which is divisible by 5
A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.