FP1 June 2013 Q9
9.
(a) A sequence of numbers is defined by \[\begin{aligned} u_1 &= 8 \\ u_{n+1} &= 4u_n - 9n, \quad n \geqslant 1 \end{aligned}\] Prove by induction that, for \(n \in \mathbb{Z}^+\), \[u_n = 4^n + 3n + 1\] (5)
(b) Prove by induction that, for \(m \in \mathbb{Z}^+\), \[\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^m = \begin{pmatrix} 2m + 1 & -4m \\ m & 1 - 2m \end{pmatrix}\] (5)
| Scheme | Marks |
|---|---|
| \(u_1 = 8\) given \(n = 1 \Rightarrow u_1 = 4^1 + 3(1) + 1 = 8\quad (\therefore \text{ true for } n = 1)\) \(4^1 + 3(1) + 1 = 8\) seen | B1 |
| Assume true for \(n = k\) so that \(u_k = 4^k + 3k + 1\) | |
| \(u_{k+1} = 4\left(4^k + 3k + 1\right) - 9k\) Substitute \(u_k\) into \(u_{k+1}\) as \(u_{k+1} = 4u_k - 9k\) | M1 |
| \(= 4^{k+1} + 12k + 4 - 9k = 4^{k+1} + 3k + 4\) Expression of the form \(4^{k+1} + ak + b\) | A1 |
| \(= 4^{k+1} + 3(k + 1) + 1\) Correct completion to an expression in terms of \(k + 1\) | A1 |
| If true for \(n = k\) then true for \(n = k + 1\) and as true for \(n = 1\) true for all \(n\) Conclusion with all 4 underlined elements that can be seen anywhere in the solution; \(n\) defined incorrectly award A0. | A1 cso |
| (5) |
| Scheme | Marks |
|---|---|
| Condone use of \(n\) here. | |
| \(lhs = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^1 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\) \(rhs = \begin{pmatrix} 2(1) + 1 & -4(1) \\ 1 & 1 - 2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\) Shows true for \(m = 1\) | B1 |
| Assume \(\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^k = \begin{pmatrix} 2k + 1 & -4k \\ k & 1 - 2k \end{pmatrix}\) | |
| \(\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^{k+1} = \begin{pmatrix} 2k + 1 & -4k \\ k & 1 - 2k \end{pmatrix}\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\) \(\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 2k + 1 & -4k \\ k & 1 - 2k \end{pmatrix}\) award M1 | M1 |
| \(= \begin{pmatrix} 6k + 3 - 4k & -8k - 4 + 4k \\ 3k + 1 - 2k & -4k - 1 + 2k \end{pmatrix}\) Or equivalent 2x2 matrix. \(\begin{pmatrix} 6k + 3 - 4k & -12k - 4 + 8k \\ 2k + 1 - k & -4k - 1 + 2k \end{pmatrix}\) award A1from above. | A1 |
| \(= \begin{pmatrix} 2k + 3 & -4k - 4 \\ k + 1 & -2k - 1 \end{pmatrix}\) | |
| \(= \begin{pmatrix} 2(k + 1) + 1 & -4(k + 1) \\ k + 1 & 1 - 2(k + 1) \end{pmatrix}\) Correct completion to a matrix in terms of \(k + 1\) | A1 |
| If true for \(m = k\) then true for \(m = k + 1\) and as true for \(m = 1\) true for all \(m\) Conclusion with all 4 underlined elements that can be seen anywhere in the solution; \(m\) defined incorrectly award A0. | A1 cso |
| (5) | |
| Total 10 |