AS October 2020 Paper 1 Q8
8. Prove by induction that, for \(n \in \mathbb{Z}^+\)
\[\mathrm{f}(n) = 2^{n+2} + 3^{2n+1}\]is divisible by 7 (6)
Way 1: \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(2^{n+2} + 3^{2n+1} = 2^3 + 3^3 = 35\) Shows the statement is true for \(n = 1\), allow 5(7) | B1 | 2.2a |
| Assume true for \(\boldsymbol{n = k}\), so \(2^{k+2} + 3^{2k+1}\) is divisible by 7 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 2^{k+3} + 3^{2k+3} - \left(2^{k+2} + 3^{2k+1}\right)\) | M1 | 2.1 |
| \(= 2 \times 2^{k+2} + 9 \times 3^{2k+1} - 2^{k+2} - 3^{2k+1}\) \(= 2^{k+2} + 8 \times 3^{2k+1}\) \(= \mathrm{f}(k) + 7 \times 3^{2k+1}\) or \(8\mathrm{f}(k) - 7 \times 2^{k+2}\) | A1 | 1.1b |
| \(\mathrm{f}(k + 1) = 2\mathrm{f}(k) + 7 \times 3^{2k+1}\) or \(9\mathrm{f}(k) - 7 \times 2^{k+2}\) | A1 | 1.1b |
| If true for \(\boldsymbol{n = k}\) then true for \(\boldsymbol{n = k + 1}\) and as it is true for \(\boldsymbol{n = 1}\) the statement is true for all (positive integers) \(\boldsymbol{n}\) | A1 | 2.4 |
| (6) | ||
| (6 marks) |
Notes
Way 1: \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)
B1: Shows that f(1) = 35 and concludes or shows divisible by 7. This may be seen in the final statement.
M1: Makes a statement that assumes the result is true for some value of \(n\)
M1: Attempts \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)
A1: Achieves a correct expression for \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) in terms of \(\mathrm{f}(k)\)
A1: Reaches a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)
A1: Correct conclusion. This mark is dependent on all previous marks, look out for scoring B1 in this statement. It is gained by conveying the ideas of all four bold 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 \(n = 1\), \(2^{n+2} + 3^{2n+1} = 2^3 + 3^3 = 35\) So the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\), so \(2^{k+2} + 3^{2k+1}\) is divisible by 7 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) = 2^{(k+1)+2} + 3^{2(k+1)+1}\) | M1 | 2.1 |
| \(\mathrm{f}(k + 1) = 2^{k+3} + 3^{2k+3} = 2 \times 2^{k+2} + 9 \times 3^{2k+1}\) \(= 2\left(2^{k+2} + 3^{2k+1}\right) + 7 \times 3^{2k+1}\) \(= 2\mathrm{f}(k) + 7 \times 3^{2k+1}\) or \(9\mathrm{f}(k) - 7 \times 2^{k+2}\) | A1 A1 | 1.1b 1.1b |
| If true for \(\boldsymbol{n = k}\) then true for \(\boldsymbol{n = k + 1}\) and as it is true for \(\boldsymbol{n = 1}\) the statement is true for all (positive integers) \(\boldsymbol{n}\) | A1 | 2.4 |
| (6) |
B1: Shows that f(1) = 35 and concludes divisible by 7
M1: Makes a statement that assumes the result is true for some value of \(n\)
M1: Attempts \(\mathrm{f}(k + 1)\)
A1: Correctly obtains either \(2\mathrm{f}(k)\) or \(7 \times 3^{2k+1}\) or either \(9\mathrm{f}(k)\) or \(-7 \times 2^{k+2}\)
A1: Reaches a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)
A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold 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 \(n = 1\), \(2^{n+2} + 3^{2n+1} = 2^3 + 3^3 = 35\) So the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\), so \(2^{k+2} + 3^{2k+1}\) is divisible by 7 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) - m\mathrm{f}(k) = 2^{k+3} + 3^{2k+3} - m\left(2^{k+2} + 3^{2k+1}\right)\) | M1 | 2.1 |
| \(= 2 \times 2^{k+2} + 9 \times 3^{2k+1} - m \times 2^{k+2} - m \times 3^{2k+1}\) \(= (2 - m)2^{k+2} + 9 \times 3^{2k+1} - m \times 3^{2k+1}\) \(= (2 - m)\left(2^{k+2} + 3^{2k+1}\right) + 7 \times 3^{2k+1}\) | A1 | 1.1b |
| \(\mathrm{f}(k + 1) = (2 - m)\left(2^{k+2} + 3^{2k+1}\right) + 7 \times 3^{2k+1} + m\mathrm{f}(k)\) | A1 | 1.1b |
| If true for \(\boldsymbol{n = k}\) then true for \(\boldsymbol{n = k + 1}\) and as it is true for \(\boldsymbol{n = 1}\) the statement is true for all (positive integers) \(\boldsymbol{n}\) | A1 | 2.4 |
| (6) |
B1: Shows that f(1) = 35 and concludes divisible by 7
M1: Makes a statement that assumes the result is true for some value of \(n\)
M1: Attempts \(\mathrm{f}(k + 1) - m\mathrm{f}(k)\)
A1: Achieves a correct expression for \(\mathrm{f}(k + 1) - m\mathrm{f}(k)\) in terms of \(\mathrm{f}(k)\)
A1: Reaches a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)
A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution.