| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove matrix power formula |
| Difficulty | Standard +0.3 This is a standard two-part induction question from FP1. Part (a) involves matrix powers with straightforward algebra, and part (b) is a routine divisibility proof. Both follow the standard induction template with no novel insights required, making this slightly easier than average for Further Maths content. |
| Spec | 4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\begin{pmatrix}2&1\\-1&0\end{pmatrix}^1 = \begin{pmatrix}2&1\\-1&0\end{pmatrix}\), for \(n=1\): \(\begin{pmatrix}n+1&n\\-n&1-n\end{pmatrix} = \begin{pmatrix}2&1\\-1&0\end{pmatrix}\) | B1 | True for \(n=1\) |
| Assume true for \(n=k\): \(\begin{pmatrix}k+1&k\\-k&1-k\end{pmatrix}\begin{pmatrix}2&1\\-1&0\end{pmatrix} = \begin{pmatrix}2k+2-k&k+1\\-2k-1+k&-k\end{pmatrix}\) | M1 A2/1/0 | |
| \(= \begin{pmatrix}(k+1)+1&k+1\\-(k+1)&1-(k+1)\end{pmatrix}\) | M1 A1 | |
| True for \(n=k+1\) if true for \(n=k\), true for \(n \in \mathbb{Z}^+\) by induction | A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(f(1) = 4 + 6 - 1 = 9 = 3\times 3\), true for \(n=1\) | B1 | |
| Assume true for \(n=k\): \(f(k) = 4^k + 6k - 1\) divisible by 3 | ||
| \(f(k+1) = 4^{k+1} + 6(k+1) - 1\) | M1 A1 | |
| \(= 4\times4^k + 6(k+1) - 1\) | A1 | |
| \(f(k+1) - f(k) = 3\times4^k + 6\) | M1 | |
| \(\therefore f(k+1) = 3(4^k+2) - f(k)\), divisible by 3 | A1 | |
| True for \(n=k+1\) if true for \(n=k\), true for \(n\in\mathbb{Z}^+\) by induction | A1 | (7 marks) |
# Question 9:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}2&1\\-1&0\end{pmatrix}^1 = \begin{pmatrix}2&1\\-1&0\end{pmatrix}$, for $n=1$: $\begin{pmatrix}n+1&n\\-n&1-n\end{pmatrix} = \begin{pmatrix}2&1\\-1&0\end{pmatrix}$ | B1 | True for $n=1$ |
| Assume true for $n=k$: $\begin{pmatrix}k+1&k\\-k&1-k\end{pmatrix}\begin{pmatrix}2&1\\-1&0\end{pmatrix} = \begin{pmatrix}2k+2-k&k+1\\-2k-1+k&-k\end{pmatrix}$ | M1 A2/1/0 | |
| $= \begin{pmatrix}(k+1)+1&k+1\\-(k+1)&1-(k+1)\end{pmatrix}$ | M1 A1 | |
| True for $n=k+1$ if true for $n=k$, true for $n \in \mathbb{Z}^+$ by induction | A1 | **(7 marks)** |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $f(1) = 4 + 6 - 1 = 9 = 3\times 3$, true for $n=1$ | B1 | |
| Assume true for $n=k$: $f(k) = 4^k + 6k - 1$ divisible by 3 | | |
| $f(k+1) = 4^{k+1} + 6(k+1) - 1$ | M1 A1 | |
| $= 4\times4^k + 6(k+1) - 1$ | A1 | |
| $f(k+1) - f(k) = 3\times4^k + 6$ | M1 | |
| $\therefore f(k+1) = 3(4^k+2) - f(k)$, divisible by 3 | A1 | |
| True for $n=k+1$ if true for $n=k$, true for $n\in\mathbb{Z}^+$ by induction | A1 | **(7 marks)** |9. Use the method of mathematical induction to prove that, for $n \in \mathbb { Z } ^ { + }$,
\begin{enumerate}[label=(\alph*)]
\item $\left( \begin{array} { c c } 2 & 1 \\ - 1 & 0 \end{array} \right) ^ { n } = \left( \begin{array} { c c } n + 1 & n \\ - n & 1 - n \end{array} \right)$
\item $\mathrm { f } ( n ) = 4 ^ { n } + 6 n - 1$ is divisible by 3 .
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 Q9 [14]}}