FP1 June 2018 Q8
8. Prove by induction that \[\mathrm{f}(n) = 2^{n+2} + 3^{2n+1}\] is divisible by 7 for all positive integers \(n\). (6)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(n) = 2^{n+2} + 3^{2n+1}\) divisible by 7 | |
| \(\mathrm{f}(1) = 2^3 + 3^3 = 35\) {which is divisible by 7}. | B1 |
| \(\{\therefore \mathrm{f}(n)\) is divisible by 7 when \(n = 1\}\) | |
| {Assume that for \(n = k\), \(\mathrm{f}(k) = 2^{k+2} + 3^{2k+1}\) is divisible by 7 for \(k \in \mathbb{Z}^+\).} | |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 2^{k+1+2} + 3^{2(k+1)+1} - (2^{k+2} + 3^{2k+1})\) | M1 |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 2(2^{k+2}) + 9(3^{2k+1}) - (2^{k+2} + 3^{2k+1})\) | |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 2^{k+2} + 8(3^{2k+1})\) | |
| \(= (2^{k+2} + 3^{2k+1}) + 7(3^{2k+1})\) or \(= 8(2^{k+2} + 3^{2k+1}) - 7(2^{k+2})\) | A1; A1 |
| \(= \mathrm{f}(k) + 7(3^{2k+1})\) or \(= 8\mathrm{f}(k) - 7(2^{k+2})\) | |
| \(\therefore \mathrm{f}(k + 1) = 2\mathrm{f}(k) + 7(3^{2k+1})\) or \(\mathrm{f}(k + 1) = 9\mathrm{f}(k) - 7(2^{k+2})\) | dM1 |
| \(\{\therefore \mathrm{f}(k + 1) = 2\mathrm{f}(k) + 7(3^{2k+1})\) is divisible by 7 as both \(2\mathrm{f}(k)\) and \(7(3^{2k+1})\) are both divisible by 7} | |
| If the result is true for \(n = k\), then it is now true for \(n = k + 1\). As the result has shown to be true for \(n = 1\), then the result is true for all \(n\) \((\in \mathbb{Z}^+)\). | A1 cso |
| (6) | |
| (6 marks) |
Notes
B1: Shows \(\mathrm{f}(1) = 35\)
M1: Applies \(\mathrm{f}(k + 1)\) with at least 1 power correct
A1; A1: \((2^{k+2} + 3^{2k+1})\) or \(\mathrm{f}(k);\ 7(3^{2k+1})\)
or \(8(2^{k+2} + 3^{2k+1})\) or \(8\mathrm{f}(k);\ -7(2^{k+2})\)
dM1: Dependent on at least one of the previous accuracy marks being awarded. Makes \(\mathrm{f}(k + 1)\) the subject
A1 cso: Correct conclusion seen at the end. Condone true for \(n = 1\) stated earlier.
ALT
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(k+1) - \alpha\mathrm{f}(k) = 2^{k+3} + 3^{2k+3} - \alpha(2^{k+2} + 3^{2k+1})\) | M1 |
| \(\mathrm{f}(k+1) - \alpha\mathrm{f}(k) = (2 - \alpha)2^{k+2} + (9 - \alpha)3^{2k+1}\) | |
| \(\mathrm{f}(k+1) - \alpha\mathrm{f}(k) = (2 - \alpha)(2^{k+2} + 3^{2k+1}) + 7\cdot 3^{2k+1}\) or \(\mathrm{f}(k+1) - \alpha\mathrm{f}(k) = (9 - \alpha)(2^{k+2} + 3^{2k+1}) - 7\cdot 2^{k+2}\) | A1;A1 |
M1: Applies \(\mathrm{f}(k + 1)\) with at least 1 power correct
A1;A1: \((2 - \alpha)(2^{k+2} + 3^{2k+1})\) or \((2 - \alpha)\mathrm{f}(k);\ 7\cdot 3^{2k+1}\) or \((9 - \alpha)(2^{k+2} + 3^{2k+1})\) or \((9 - \alpha)\mathrm{f}(k);\ -7\cdot 2^{k+2}\)
NB: Choosing \(\alpha = 0, \alpha = 2, \alpha = 9\) will make relevant terms disappear, but marks should be awarded accordingly.