| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2019 |
| 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 straightforward proof by induction for divisibility. The base case is trivial (n=1 gives 3^6 - 2^2 = 729-4=725=5×145). The inductive step requires standard algebraic manipulation: expressing f(k+1) in terms of f(k) by factoring out common terms (3^2 and 2^2), then showing the result is divisible by 5. While it requires careful algebra, this is a textbook application of induction with no novel insight needed, making it slightly easier than average. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| When \(n=1\): \(3^{2n+4} - 2^{2n} = 729 - 4 = 725 = 145 \times 5\), so true for \(n=1\) | B1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Assume true for \(n=k\) so \(3^{2k+4} - 2^{2k}\) is divisible by 5 | M1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(k+1) - f(k) = 3^{2k+6} - 2^{2k+2} - 3^{2k+4} + 2^{2k}\) | M1 | 2.1 |
| either \(8f(k) + 5 \times 2^{2k}\) or \(3f(k) + 5 \times 3^{2k+4}\) | A1 | 1.1b |
| \(f(k+1) = 9f(k) + 5 \times 2^{2k}\) or \(f(k+1) = 4f(k) + 5 \times 3^{2k+4}\) | A1 | 1.1b |
| If true for \(n=k\) then true for \(n=k+1\); true for \(n=1\), therefore true for all positive integers \(n\) | A1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(k+1) = 3^{2(k+1)+4} - 2^{2(k+1)} = 3^{2k+6} - 2^{2k+2}\) | M1 | 2.1 |
| \(f(k+1) = 9f(k) + 5 \times 2^{2k}\) or \(f(k+1) = 4f(k) + 5 \times 3^{2k+4}\) | A1, A1 | 1.1b, 1.1b |
| Conclusion as above | A1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(k+1) = 3^{2(k+1)+4} - 2^{2(k+1)} = 3^{2k+6} - 2^{2k+2}\) | M1 | 2.1 |
| \(f(k+1) = 45M + 5 \times 2^{2k}\) or \(f(k+1) = 5 \times 3^{2k+4} + 20M\) | A1, A1 | 1.1b, 1.1b |
| Conclusion as above | A1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(k+1) + f(k) = 3^{2k+6} - 2^{2k+2} + 3^{2k+4} - 2^{2k}\) | M1 | 2.1 |
| Leading to \(10 \times 3^{2k+4} - 5 \times 2^{2k}\) | ||
| \(f(k+1) = 5\left[2 \times 3^{2k+4} - 2^{2k}\right] - f(k)\) | A1, A1 | 1.1b, 1.1b |
| Conclusion as above | A1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(k+1) - Mf(k) = 3^{2k+6} - 2^{2k+2} - M \times 3^{2k+4} + M \times 2^{2k}\) | M1 | 2.1 |
| \(f(k+1) - Mf(k) = (9-M) \times 3^{2k+4} - (4-M) \times 2^{2k}\) | A1 | 1.1b |
| \(f(k+1) = (9-M) \times 3^{2k+4} - (4-M) \times 2^{2k} + Mf(k)\) | A1 | 1.1b |
| Conclusion as above | A1 | 2.4 |
# Question 6:
**Proof by induction that $3^{2n+4} - 2^{2n}$ is divisible by 5**
## Base Case
| When $n=1$: $3^{2n+4} - 2^{2n} = 729 - 4 = 725 = 145 \times 5$, so true for $n=1$ | B1 | 2.2a | Must show f(1) = 725 and conclusion true for $n=1$ |
## Inductive Step - Ways 1 & 2
| Assume true for $n=k$ so $3^{2k+4} - 2^{2k}$ is divisible by 5 | M1 | 2.4 | Assumption statement; may be recovered in conclusion |
**Way 1** $f(k+1) - f(k)$:
| $f(k+1) - f(k) = 3^{2k+6} - 2^{2k+2} - 3^{2k+4} + 2^{2k}$ | M1 | 2.1 | Attempts $f(k+1) - f(k)$ |
| either $8f(k) + 5 \times 2^{2k}$ or $3f(k) + 5 \times 3^{2k+4}$ | A1 | 1.1b | Correct simplified expression |
| $f(k+1) = 9f(k) + 5 \times 2^{2k}$ or $f(k+1) = 4f(k) + 5 \times 3^{2k+4}$ | A1 | 1.1b | Correct expression for $f(k+1)$ in terms of $f(k)$ |
| If true for $n=k$ then true for $n=k+1$; true for $n=1$, therefore true for all positive integers $n$ | A1 | 2.4 | Dependent on **all** previous marks; all underlined points conveyed |
**Way 2** $f(k+1)$:
| $f(k+1) = 3^{2(k+1)+4} - 2^{2(k+1)} = 3^{2k+6} - 2^{2k+2}$ | M1 | 2.1 | Attempts $f(k+1)$ |
| $f(k+1) = 9f(k) + 5 \times 2^{2k}$ or $f(k+1) = 4f(k) + 5 \times 3^{2k+4}$ | A1, A1 | 1.1b, 1.1b | Correct simplified; correct in terms of $f(k)$ |
| Conclusion as above | A1 | 2.4 | |
**Way 3** $f(k) = 5M$:
| $f(k+1) = 3^{2(k+1)+4} - 2^{2(k+1)} = 3^{2k+6} - 2^{2k+2}$ | M1 | 2.1 | |
| $f(k+1) = 45M + 5 \times 2^{2k}$ or $f(k+1) = 5 \times 3^{2k+4} + 20M$ | A1, A1 | 1.1b, 1.1b | Correctly achieves $45M$ or $5\times2^{2k}$ or $20M$ or $5\times3^{2k+4}$ |
| Conclusion as above | A1 | 2.4 | |
**Way 4** $f(k+1) + f(k)$:
| $f(k+1) + f(k) = 3^{2k+6} - 2^{2k+2} + 3^{2k+4} - 2^{2k}$ | M1 | 2.1 | |
| Leading to $10 \times 3^{2k+4} - 5 \times 2^{2k}$ | | | |
| $f(k+1) = 5\left[2 \times 3^{2k+4} - 2^{2k}\right] - f(k)$ | A1, A1 | 1.1b, 1.1b | |
| Conclusion as above | A1 | 2.4 | |
**Way 5** $f(k+1) - Mf(k)$:
| $f(k+1) - Mf(k) = 3^{2k+6} - 2^{2k+2} - M \times 3^{2k+4} + M \times 2^{2k}$ | M1 | 2.1 | |
| $f(k+1) - Mf(k) = (9-M) \times 3^{2k+4} - (4-M) \times 2^{2k}$ | A1 | 1.1b | |
| $f(k+1) = (9-M) \times 3^{2k+4} - (4-M) \times 2^{2k} + Mf(k)$ | A1 | 1.1b | Correct expression divisible by 5 |
| Conclusion as above | A1 | 2.4 | |
---\begin{enumerate}
\item Prove by induction that for all positive integers $n$
\end{enumerate}
$$f ( n ) = 3 ^ { 2 n + 4 } - 2 ^ { 2 n }$$
is divisible by 5\\
(6)
\hfill \mbox{\textit{Edexcel CP1 2019 Q6 [6]}}