| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove divisibility |
| Difficulty | Standard +0.3 This is a standard proof by induction for divisibility with straightforward algebra. The base case is trivial (n=1 gives 8+27=35=7×5), and the inductive step requires routine manipulation: expressing f(k+1) in terms of f(k) and factoring out 7. While it's a Further Maths topic, it follows a well-established template with no conceptual surprises, making it slightly easier than average overall. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(n) = 2^{n+2} + 3^{2n+1}\) divisible by 7 | ||
| \(f(1) = 2^3 + 3^3 = 35\) {which is divisible by 7}. | B1 | Shows \(f(1) = 35\) |
| \(\{∴f(n)\) is divisible by 7 when \(n = 1\}\) | ||
| \(\{\)Assume that for \(n = k\), \(f(k) = 2^{k+2} + 3^{2k+1}\) is divisible by 7 for \(k \in \mathbb{Z}^+.\}\) | ||
| \(f(k+1) - f(k) = 2^{k+1+2} + 3^{2(k+1)+1} - (2^{k+2} + 3^{2k+1})\) | M1 | Applies \(f(k+1)\) with at least 1 power correct |
| \(f(k+1) - f(k) = 2(2^{k+2}) + 9(3^{2k+1}) - (2^{k+2} + 3^{2k+1})\) | ||
| \(f(k+1) - f(k) = 2^{k+2} + 8(3^{2k+1}) = (2^{k+2} + 3^{2k+1}) + 7(3^{2k+1})\) | A1; A1 | \((2^{k+2} + 3^{2k+1})\) or \(f(k)\); \(7(3^{2k+1})\) |
| \(\text{or} = 8(2^{k+2}) + 3^{2k+1}) - 7(2^{k+2})\) | or \(8(2^{k+2} + 3^{2k+1})\) or \(8f(k); -7(2^{k+2})\) | |
| \(\text{or} = 8f(k) - 7(2^{k+2})\) | ||
| \(∴f(k+1) = 2f(k) + 7(3^{2k+1})\) or \(f(k+1) = 9f(k) - 7(2^{k+2})\) | dM1 | Dependent on at least one of the previous accuracy marks being awarded. Makes \(f(k+1)\) the subject |
| \(\{∴f(k+1) = 2f(k) + 7(3^{2k+1})\) is divisible by 7 as both \(2f(k)\) and \(7(3^{2k+1})\) are both divisible by 7\}$ | ||
| If the result is true for \(n = k\), then it is now true for \(n = k+1\). As the result has shown to be true for \(n = 1\), then the result is true for all \(n (\in \mathbb{Z}^+).\) | A1 cso | Correct conclusion seen at the end. Condone true for \(n = 1\) stated earlier. |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(k+1) - \alpha f(k) = 2^{k+3} + 3^{2k+3} - \alpha(2^{k+2} + 3^{2k+1})\) | M1 | Applies \(f(k+1)\) with at least 1 power correct |
| \(f(k+1) - \alpha f(k) = (2-\alpha)2^{k+2} + (9-\alpha)3^{2k+1}\) | ||
| \(f(k+1) - \alpha f(k) = (2-\alpha)2^{k+2} + 3^{2k+1}) - 7.2^{k+2}\) or \((9-\alpha)(2^{k+2} + 3^{2k+1}) - 7.2^{k+2}\) | A1;A1 | \((2-\alpha)(2^{k+2} + 3^{2k+1})\) or \((2-\alpha)f(k); -7(2^{k+2})\) or \((9-\alpha)(2^{k+2} + 3^{2k+1})\) or \((9-\alpha)f(k); -7.2^{k+2}\) |
| NB: Choosing \(\alpha = 0, \alpha = 2, \alpha = 9\) will make relevant terms disappear, but marks should be awarded accordingly. |
| $f(n) = 2^{n+2} + 3^{2n+1}$ divisible by 7 | | |
| $f(1) = 2^3 + 3^3 = 35$ {which is divisible by 7}. | B1 | Shows $f(1) = 35$ |
| $\{∴f(n)$ is divisible by 7 when $n = 1\}$ | | |
| $\{$Assume that for $n = k$, $f(k) = 2^{k+2} + 3^{2k+1}$ is divisible by 7 for $k \in \mathbb{Z}^+.\}$ | | |
| $f(k+1) - f(k) = 2^{k+1+2} + 3^{2(k+1)+1} - (2^{k+2} + 3^{2k+1})$ | M1 | Applies $f(k+1)$ with at least 1 power correct |
| $f(k+1) - f(k) = 2(2^{k+2}) + 9(3^{2k+1}) - (2^{k+2} + 3^{2k+1})$ | | |
| $f(k+1) - f(k) = 2^{k+2} + 8(3^{2k+1}) = (2^{k+2} + 3^{2k+1}) + 7(3^{2k+1})$ | A1; A1 | $(2^{k+2} + 3^{2k+1})$ or $f(k)$; $7(3^{2k+1})$ |
| $\text{or} = 8(2^{k+2}) + 3^{2k+1}) - 7(2^{k+2})$ | | or $8(2^{k+2} + 3^{2k+1})$ or $8f(k); -7(2^{k+2})$ |
| $\text{or} = 8f(k) - 7(2^{k+2})$ | | |
| $∴f(k+1) = 2f(k) + 7(3^{2k+1})$ or $f(k+1) = 9f(k) - 7(2^{k+2})$ | dM1 | Dependent on at least one of the previous accuracy marks being awarded. Makes $f(k+1)$ the subject |
| $\{∴f(k+1) = 2f(k) + 7(3^{2k+1})$ is divisible by 7 as both $2f(k)$ and $7(3^{2k+1})$ are both divisible by 7\}$ | | |
| If the result is true for $n = k$, then it is now true for $n = k+1$. As the result has shown to be true for $n = 1$, then the result is true for all $n (\in \mathbb{Z}^+).$ | A1 cso | Correct conclusion seen at the end. Condone true for $n = 1$ stated earlier. |
**ALT**
| $f(k+1) - \alpha f(k) = 2^{k+3} + 3^{2k+3} - \alpha(2^{k+2} + 3^{2k+1})$ | M1 | Applies $f(k+1)$ with at least 1 power correct |
| $f(k+1) - \alpha f(k) = (2-\alpha)2^{k+2} + (9-\alpha)3^{2k+1}$ | | |
| $f(k+1) - \alpha f(k) = (2-\alpha)2^{k+2} + 3^{2k+1}) - 7.2^{k+2}$ or $(9-\alpha)(2^{k+2} + 3^{2k+1}) - 7.2^{k+2}$ | A1;A1 | $(2-\alpha)(2^{k+2} + 3^{2k+1})$ or $(2-\alpha)f(k); -7(2^{k+2})$ or $(9-\alpha)(2^{k+2} + 3^{2k+1})$ or $(9-\alpha)f(k); -7.2^{k+2}$ |
| | | NB: Choosing $\alpha = 0, \alpha = 2, \alpha = 9$ will make relevant terms disappear, but marks should be awarded accordingly. |
**Total: 6 marks**
---\begin{enumerate}
\item Prove by induction that
\end{enumerate}
$$f ( n ) = 2 ^ { n + 2 } + 3 ^ { 2 n + 1 }$$
is divisible by 7 for all positive integers $n$.\\
\hfill \mbox{\textit{Edexcel FP1 2018 Q8 [6]}}