| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove divisibility |
| Difficulty | Standard +0.3 This is a straightforward two-part induction question with standard techniques. Part (a) is routine divisibility proof requiring basic modular arithmetic (5^n ≡ 1 mod 4). Part (b) is matrix induction with simple algebra. Both follow textbook templates with no novel insights needed, making this slightly easier than average despite being Further Maths content. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(f(1) = 5 + 8 + 3 = 16\), (which is divisible by 4). (\(\therefore\) True for \(n=1\)) | B1 | Correct values of 16 or 4 for \(n=1\) or \(n=0\) (accept "is a multiple of") |
| Using the formula to write down \(f(k+1)\): \(f(k+1) = 5^{k+1} + 8(k+1) + 3\) | M1, A1 | M1 using the formula to write down \(f(k+1)\); A1 correct expression |
| \(f(k+1) - f(k) = 5^{k+1} + 8(k+1) + 3 - 5^k - 8k - 3\) \(= 5(5^k) + 8k + 8 + 3 - 5^k - 8k - 3 = 4(5^k) + 8\) | M1, A1 | M1 start method to connect \(f(k+1)\) with \(f(k)\) |
| \(f(k+1) = 4(5^k + 2) + f(k)\), which is divisible by 4 | A1ft | A1 correct working toward multiples of 4, with \(f(k+1)\) as subject |
| \(\therefore\) True for \(n = k+1\) if true for \(n = k\). True for \(n=1\), \(\therefore\) true for all \(n\) | A1cso | Full induction conclusion required |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(f(k+1) = 5(5^k) + 8k + 8 + 3 = 4(5^k) + 8 + (5^k + 8k + 3) = 4(5^k+2) + f(k)\) | M1, A1 | Or \(= 5f(k) - 4(8k+1)\), which is divisible by 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| 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 | Correct statement for \(n=1\) or \(n=0\) |
| \(\begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix}^{k+1} = \begin{pmatrix} 2k+1 & -2k \\ 2k & 1-2k \end{pmatrix}\begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 2k+3 & -2k-2 \\ 2k+2 & -2k-1 \end{pmatrix}\) | M1, A1, A1 | M1 set up product of two appropriate matrices; A1A0 for one or two slips; A1A1 all correct simplified; A0A0 more than two slips |
| \(= \begin{pmatrix} 2(k+1)+1 & -2(k+1) \\ 2(k+1) & 1-2(k+1) \end{pmatrix}\) | M1, A1 | M1 states in terms of \((k+1)\); A1 correct statement |
| \(\therefore\) True for \(n = k+1\) if true for \(n = k\). True for \(n=1\), \(\therefore\) true for all \(n\) | A1cso | Full induction conclusion required |
# Question 8:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $f(1) = 5 + 8 + 3 = 16$, (which is divisible by 4). ($\therefore$ True for $n=1$) | B1 | Correct values of 16 or 4 for $n=1$ or $n=0$ (accept "is a multiple of") |
| Using the formula to write down $f(k+1)$: $f(k+1) = 5^{k+1} + 8(k+1) + 3$ | M1, A1 | M1 using the formula to write down $f(k+1)$; A1 correct expression |
| $f(k+1) - f(k) = 5^{k+1} + 8(k+1) + 3 - 5^k - 8k - 3$ $= 5(5^k) + 8k + 8 + 3 - 5^k - 8k - 3 = 4(5^k) + 8$ | M1, A1 | M1 start method to connect $f(k+1)$ with $f(k)$ |
| $f(k+1) = 4(5^k + 2) + f(k)$, **which is divisible by 4** | A1ft | A1 correct working toward multiples of 4, with $f(k+1)$ as subject |
| $\therefore$ True for $n = k+1$ **if** true for $n = k$. True for $n=1$, $\therefore$ true for all $n$ | A1cso | Full induction conclusion required |
**Alternative for 2nd M:**
| Working | Marks | Guidance |
|---------|-------|----------|
| $f(k+1) = 5(5^k) + 8k + 8 + 3 = 4(5^k) + 8 + (5^k + 8k + 3) = 4(5^k+2) + f(k)$ | M1, A1 | Or $= 5f(k) - 4(8k+1)$, which is divisible by 4 |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| 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 | Correct statement for $n=1$ or $n=0$ |
| $\begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix}^{k+1} = \begin{pmatrix} 2k+1 & -2k \\ 2k & 1-2k \end{pmatrix}\begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 2k+3 & -2k-2 \\ 2k+2 & -2k-1 \end{pmatrix}$ | M1, A1, A1 | M1 set up product of two appropriate matrices; A1A0 for one or two slips; A1A1 all correct simplified; A0A0 more than two slips |
| $= \begin{pmatrix} 2(k+1)+1 & -2(k+1) \\ 2(k+1) & 1-2(k+1) \end{pmatrix}$ | M1, A1 | M1 states in terms of $(k+1)$; A1 correct statement |
| $\therefore$ True for $n = k+1$ **if** true for $n = k$. **True for $n=1$**, $\therefore$ **true for all $n$** | A1cso | Full induction conclusion required |8. Prove by induction that, for $n \in \mathbb { Z } ^ { + }$,
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { f } ( n ) = 5 ^ { n } + 8 n + 3$ is divisible by 4 ,
\item $\left( \begin{array} { l l } 3 & - 2 \\ 2 & - 1 \end{array} \right) ^ { n } = \left( \begin{array} { l r } 2 n + 1 & - 2 n \\ 2 n & 1 - 2 n \end{array} \right)$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2009 Q8 [14]}}