FP1 June 2014 Q9
9. Prove by induction that, for \(n \in \mathbb{Z}^{+}\), \[\mathrm{f}(n) = 8^n - 2^n\] is divisible by 6 (6)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(n) = 8^n - 2^n\) is divisible by 6. | |
| \(\mathrm{f}(1) = 8^1 - 2^1 = 6\), Shows that \(\mathrm{f}(1) = 6\) | B1 |
| Assume that for \(n = k\), \(\mathrm{f}(k) = 8^k - 2^k\) is divisible by 6. | |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 8^{k+1} - 2^{k+1} - (8^k - 2^k)\) Attempt \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) | M1 |
| \(= 8^k(8 - 1) + 2^k(1 - 2) = 7 \times 8^k - 2^k\) | |
| \(= 6 \times 8^k + 8^k - 2^k\ \big(= 6 \times 8^k + \mathrm{f}(k)\big)\) M1: Attempt \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) as a multiple of 6 A1: rhs a correct multiple of 6 | M1A1 |
| \(\mathrm{f}(k + 1) = 6 \times 8^k + 2\mathrm{f}(k)\) Completes to \(\mathrm{f}(k + 1) =\) a multiple of 6 | A1 |
| If the result is true for \(n = k\), then it is now true for \(n = k + 1\). As the result has been shown to be true for \(n = 1\), then the result is true for all \(n\ (\in \mathbb{Z}^{+}.)\) | A1cso |
Do not award final A if \(n\) defined incorrctly e.g. ‘\(n\) is an integer’ award A0 | |
| (6) | |
| (6 marks) |
Alternative methods
Way 2
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1) = 8^1 - 2^1 = 6\), Shows that \(\mathrm{f}(1) = 6\) | B1 |
| Assume that for \(n = k\), \(\mathrm{f}(k) = 8^k - 2^k\) is divisible by 6. | |
| \(\mathrm{f}(k + 1) = 8^{k+1} - 2^{k+1} = 8(8^k - 2^k + 2^k) - 2.2^k\) Attempts \(\mathrm{f}(k + 1)\) in terms of \(2^k\) and \(8^k\) | M1 |
| \(\mathrm{f}(k + 1) = 8^{k+1} - 2^{k+1} = 8(\mathrm{f}(k) + 2^k) - 2.2^k\) M1: Attempts \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\) A1: rhs correct and a multiple of 6 | M1A1 |
| \(\mathrm{f}(k + 1) = 8\mathrm{f}(k) + 6.2^k\) Completes to \(\mathrm{f}(k + 1) =\) a multiple of 6 | A1 |
| If the result is true for \(n = k\), then it is now true for \(n = k + 1\). As the result has been shown to be true for \(n = 1\), then the result is true for all \(n\ (\in \mathbb{Z}^{+}.)\) | A1cso |
Way 3
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1) = 8^1 - 2^1 = 6\), Shows that \(\mathrm{f}(1) = 6\) | B1 |
| Assume that for \(n = k\), \(\mathrm{f}(k) = 8^k - 2^k\) is divisible by 6. | |
| \(\mathrm{f}(k + 1) - 8\mathrm{f}(k) = 8^{k+1} - 2^{k+1} - 8.8^k + 8.2^k\) Attempt \(\mathrm{f}(k + 1) - 8\mathrm{f}(k)\) Any multiple \(m\) replacing 8 award M1 | M1 |
| \(\mathrm{f}(k + 1) - 8\mathrm{f}(k) = 8^{k+1} - 8^{k+1} + 8.2^k - 2.2^k = 6.2^k\) M1: Attempt \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) as a multiple of 6 A1: rhs a correct multiple of 6 | M1A1 |
| \(\mathrm{f}(k + 1) = 8\mathrm{f}(k) + 6.2^k\) Completes to \(\mathrm{f}(k + 1) =\) a multiple of 6 | A1 |
General Form for multiple \(m\) \(\mathrm{f}(k + 1) - m\mathrm{f}(k) = 6.8^k + (2 - m)(8^k - 2^k)\) (corrected from the printed mark scheme: \(\mathrm{f}(k + 1) = 6.8^k + (2 - m)(8^k - 2^k)\)) | |
| If the result is true for \(n = k\), then it is now true for \(n = k + 1\). As the result has been shown to be true for \(n = 1\), then the result is true for all \(n\ (\in \mathbb{Z}^{+}.)\) | A1cso |