FP1 June 2009 Q8
8. Prove by induction that, for \(n \in \mathbb{Z}^+\),
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1) = 5 + 8 + 3 = 16\), (which is divisible by 4). (\(\therefore\) True for \(n = 1\)). | B1 |
| Using the formula to write down \(\mathrm{f}(k + 1)\), \(\mathrm{f}(k + 1) = 5^{k+1} + 8(k + 1) + 3\) | M1 A1 |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 5^{k+1} + 8(k + 1) + 3 - 5^k - 8k - 3\) | M1 |
| \(= 5(5^k) + 8k + 8 + 3 - 5^k - 8k - 3 = 4(5^k) + 8\) | A1 |
| \(\mathrm{f}(k + 1) = 4(5^k + 2) + \mathrm{f}(k)\), which is divisible by 4 | A1ft |
| \(\therefore\) True for \(n = k + 1\) if true for \(n = k\). True for \(n = 1\), \(\therefore\) true for all \(n\). | A1cso |
| (7) |
(a) Alternative for 2nd M:
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(k + 1) = 5(5^k) + 8k + 8 + 3\) | M1 |
| \(= 4(5^k) + 8 + (5^k + 8k + 3)\) or \(= 5(5^k + 8k + 3) - 32k - 4\) | A1 |
| \(= 4(5^k + 2) + \mathrm{f}(k)\), or \(= 5\mathrm{f}(k) - 4(8k + 1)\) which is divisible by 4 | A1 (or similar methods) |
Notes
(a) B1 Correct values of 16 or 4 for \(n = 1\) or for \(n = 0\) (Accept “is a multiple of”)
M1 Using the formula to write down \(\mathrm{f}(k + 1)\) A1 Correct expression of \(\mathrm{f}(k+1)\) (or for \(\mathrm{f}(n + 1)\))
M1 Start method to connect \(\mathrm{f}(k+1)\) with \(\mathrm{f}(k)\) as shown
A1 correct working toward multiples of 4, A1 ft result including \(\mathrm{f}(k + 1)\) as subject, A1cso conclusion
| Scheme | Marks |
|---|---|
| For \(n = 1\), \(\begin{pmatrix} 2n + 1 & -2n \\ 2n & 1 - 2n \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix}^1\) (\(\therefore\) True for \(n = 1\).) | B1 |
| \(\begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix}^{k+1} = \begin{pmatrix} 2k + 1 & -2k \\ 2k & 1 - 2k \end{pmatrix}\begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 2k + 3 & -2k - 2 \\ 2k + 2 & -2k - 1 \end{pmatrix}\) | M1 A1 A1 |
| \(= \begin{pmatrix} 2(k + 1) + 1 & -2(k + 1) \\ 2(k + 1) & 1 - 2(k + 1) \end{pmatrix}\) | M1 A1 |
| \(\therefore\) True for \(n = k + 1\) if true for \(n = k\). True for \(\boldsymbol{n} = 1\), \(\therefore\) true for all \(\boldsymbol{n}\) | A1 cso |
| (7) | |
| [14] |
Notes
(b) B1 correct statement for \(n = 1\) or \(n = 0\)
First M1: Set up product of two appropriate matrices – product can be either way round
A1 A0 for one or two slips in simplified result
A1 A1 all correct simplified
A0 A0 more than two slips
M1: States in terms of (k + 1)
A1 Correct statement A1 for induction conclusion
Part (b) Alternative
May write \(\begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix}^{k+1} = \begin{pmatrix} 2k + 3 & -2k - 2 \\ 2k + 2 & -2k - 1 \end{pmatrix}\). Then may or may not complete the proof.
This can be awarded the second M (substituting \(k + 1\)) and following A (simplification) in part (b). The first three marks are awarded as before. Concluding that they have reached the same matrix and therefore a result will then be part of final A1 cso but also need other statements as in the first method.