| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove recurrence relation formula |
| Difficulty | Standard +0.3 This is a standard FP1 induction question with routine structure: prove a recurrence relation formula and a matrix power formula, both following textbook templates. Part (b)(ii) adds mild interest by testing n=-1, but overall this requires only methodical application of the induction algorithm with straightforward algebra—slightly easier than the average A-level question due to its predictable structure and lack of conceptual obstacles. |
| Spec | 4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(n=1\): \(u_1 = 3 + 2(1-1) = 3\), so result true for \(n=1\) | B1 | |
| Assume true for \(k\); Then \(u_{k+1} = 3(3^k + 2(3^{k-1}-1))+4\) | M1 | |
| So \(u_{k+1} = 3^{k+1} + 2(3^k) - 6 + 4\) | M1 | |
| \(u_{k+1} = 3^{k+1} + 2(3^k) - 2 = 3^{k+1} + 2(3^k - 1)\), so result true for \(k+1\) | A1 | |
| So by induction result is true for all positive integers | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{A}^1 = \begin{pmatrix} 4 & 0 \\ 3\times3 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 9 & 1 \end{pmatrix}\) so result true for \(n=1\) | B1 | |
| Then \(\mathbf{A}^{k+1} = \begin{pmatrix} 4 & 0 \\ 9 & 1 \end{pmatrix}\begin{pmatrix} 4^k & 0 \\ 3(4^k-1) & 1 \end{pmatrix}\) | B1 | |
| \(= \begin{pmatrix} 4^{k+1} & 0 \\ 9\cdot4^k + 3\cdot4^k - 3 & 1 \end{pmatrix}\) | M1 | |
| \(= \begin{pmatrix} 4^{k+1} & 0 \\ 3\cdot4^k(3+1)-3 & 1 \end{pmatrix} = \begin{pmatrix} 4^{k+1} & 0 \\ 3\cdot4^{k+1}-3 & 1 \end{pmatrix}\) | M1 A1 | |
| So result true for \(k+1\) | M1 | |
| So by induction result is true for all positive integers | A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(n=-1\), \(\mathbf{A}^{-1} = \begin{pmatrix} \frac{1}{4} & 0 \\ -\frac{9}{4} & 1 \end{pmatrix}\) | M1 | |
| This is the correct inverse of \(\mathbf{A}\), so result is valid | A1 | (2 marks) |
| (14 marks total) |
# Question 9:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $n=1$: $u_1 = 3 + 2(1-1) = 3$, so result true for $n=1$ | B1 | |
| Assume true for $k$; Then $u_{k+1} = 3(3^k + 2(3^{k-1}-1))+4$ | M1 | |
| So $u_{k+1} = 3^{k+1} + 2(3^k) - 6 + 4$ | M1 | |
| $u_{k+1} = 3^{k+1} + 2(3^k) - 2 = 3^{k+1} + 2(3^k - 1)$, so result true for $k+1$ | A1 | |
| So by induction result is true for all positive integers | A1 | (5 marks) |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{A}^1 = \begin{pmatrix} 4 & 0 \\ 3\times3 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 9 & 1 \end{pmatrix}$ so result true for $n=1$ | B1 | |
| Then $\mathbf{A}^{k+1} = \begin{pmatrix} 4 & 0 \\ 9 & 1 \end{pmatrix}\begin{pmatrix} 4^k & 0 \\ 3(4^k-1) & 1 \end{pmatrix}$ | B1 | |
| $= \begin{pmatrix} 4^{k+1} & 0 \\ 9\cdot4^k + 3\cdot4^k - 3 & 1 \end{pmatrix}$ | M1 | |
| $= \begin{pmatrix} 4^{k+1} & 0 \\ 3\cdot4^k(3+1)-3 & 1 \end{pmatrix} = \begin{pmatrix} 4^{k+1} & 0 \\ 3\cdot4^{k+1}-3 & 1 \end{pmatrix}$ | M1 A1 | |
| So result true for $k+1$ | M1 | |
| So by induction result is true for all positive integers | A1 | (7 marks) |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $n=-1$, $\mathbf{A}^{-1} = \begin{pmatrix} \frac{1}{4} & 0 \\ -\frac{9}{4} & 1 \end{pmatrix}$ | M1 | |
| This is the correct inverse of $\mathbf{A}$, so result is valid | A1 | (2 marks) |
| **(14 marks total)** | | |9. (a) A sequence of numbers is defined by
$$u _ { 1 } = 3 \text { and } u _ { n + 1 } = 3 u _ { n } + 4 \text { for } n \geqslant 1 .$$
Prove by induction that
$$u _ { n } = 3 ^ { n } + 2 \left( 3 ^ { n - 1 } - 1 \right) \text { for } n \in \mathbb { Z } ^ { + } \text {. }$$
(b)
$$\mathbf { A } = \left( \begin{array} { l l }
4 & 0 \\
9 & 1
\end{array} \right)$$
\begin{enumerate}[label=(\roman*)]
\item Prove by induction that
$$\mathbf { A } ^ { n } = \left( \begin{array} { c c }
4 ^ { n } & 0 \\
3 \left( 4 ^ { n } - 1 \right) & 1
\end{array} \right) \text { for } n \in \mathbb { Z } ^ { + } .$$
\item Determine whether the result $\mathbf { A } ^ { n } = \left( \begin{array} { c c } 4 ^ { n } & 0 \\ 3 \left( 4 ^ { n } - 1 \right) & 1 \end{array} \right)$ is also valid for $n = - 1$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 Q9 [14]}}