FP1 June 2010 Q7
7. \[\mathrm{f}(n) = 2^n + 6^n\]
| Scheme | Marks |
|---|---|
| LHS \(= \mathrm{f}(k + 1) = 2^{k+1} + 6^{k+1}\) | M1 |
| \(= 2(2^k) + 6(6^k)\) | A1 |
| \(= 6(2^k + 6^k) - 4(2^k) = 6\mathrm{f}(k) - 4(2^k)\) | A1 |
| (3) |
OR
| Scheme | Marks |
|---|---|
| RHS \(= 6\mathrm{f}(k) - 4(2^k) = 6(2^k + 6^k) - 4(2^k)\) | M1 |
| \(= 2(2^k) + 6(6^k)\) | A1 |
| \(= 2^{k+1} + 6^{k+1} = \mathrm{f}(k + 1)\) (*) | A1 |
| (3) |
OR
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(k + 1) - 6\mathrm{f}(k) = 2^{k+1} + 6^{k+1} - 6(2^k + 6^k)\) | M1 |
| \(= (2 - 6)(2^k) = -4 \cdot 2^k\), and so \(\mathrm{f}(k + 1) = 6\mathrm{f}(k) - 4(2^k)\) | A1, A1 |
| (3) |
Notes
(a) M1: for substitution into LHS (or RHS) or \(\mathrm{f}(k + 1) - 6\mathrm{f}(k)\)
A1: for correct split of the two separate powers cao
A1: for completion of proof with no error or ambiguity (needs (for example) to start with one side of equation and reach the other or show that each side separately is \(2(2^k) + 6(6^k)\) and conclude LHS = RHS)
| Scheme | Marks |
|---|---|
| \(n = 1\): \(\mathrm{f}(1) = 2^1 + 6^1 = 8\), which is divisible by 8 | B1 |
| Either Assume f(\(k\)) divisible by 8 and try to use \(\mathrm{f}(k + 1) = 6\mathrm{f}(k) - 4(2^k)\) | M1 |
| Show \(4(2^k) = 4 \times 2(2^{k-1}) = 8(2^{k-1})\) or \(8\left(\tfrac{1}{2}2^k\right)\) Or valid statement | A1 |
| Deduction that result is implied for \(n = k + 1\) and so is true for positive integers by induction (may include \(n = 1\) true here) | A1cso |
| (4) | |
| 7 marks |
Or
| Scheme | Marks |
|---|---|
| Or Assume f(\(k\)) divisible by 8 and try to use \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) or \(\mathrm{f}(k + 1) + \mathrm{f}(k)\) including factorising \(6^k = 2^k3^k\) | M1 |
| \(= 2^3 2^{k-3}(1 + 5 \cdot 3^k)\) or \(= 2^3 2^{k-3}(3 + 7 \cdot 3^k)\) o.e. | A1 |
| Deduction that result is implied for \(n = k + 1\) and so is true for positive integers by induction (must include explanation of why \(n = 2\) is also true here) | A1cso |
Notes
(b) B1: for substitution of \(n = 1\) and stating “true for \(n = 1\)” or “divisible by 8” or tick. (This statement may appear in the concluding statement of the proof)
M1: Assume f(\(k\)) divisible by 8 and consider \(\mathrm{f}(k + 1) = 6\mathrm{f}(k) - 4(2^k)\) or equivalent expression that could lead to proof – not merely \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) unless deduce that 2 is a factor of 6 (see right hand scheme above).
A1: Indicates each term divisible by 8 OR takes out factor 8 or \(2^3\)
A1: Induction statement. Statement \(n = 1\) here could contribute to B1 mark earlier.
NB: \(f(k + 1) - f(k) = 2^{k+1} - 2^k + 6^{k+1} - 6^k = 2^k + 5 \cdot 6^k\) only is M0 A0 A0
(b) “Otherwise” methods
Could use: \(\mathrm{f}(k + 1) = 2\mathrm{f}(k) + 4(6^k)\) or \(\mathrm{f}(k + 2) = 36\mathrm{f}(k) - 32(6^k)\) or \(\mathrm{f}(k + 2) = 4\mathrm{f}(k) + 32(2^k)\) in a similar way to given expression and Left hand mark scheme is applied.
Special Case: Otherwise Proof not involving induction: This can only be awarded the B1 for checking \(n = 1\).