| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Session | Specimen |
| 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, requiring students to verify the base case, assume for n=k, and prove for n=k+1 by manipulating the expression to show it contains a factor of 17. The algebraic manipulation is straightforward once students recognize they need to express f(k+1) in terms of f(k), making this slightly easier than average but still requiring proper induction technique. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(n=1\), \(2^{3n+1}+3(5^{2n+1})=16+375=391\); \(391=17\times23\) so true for \(n=1\) | B1 | Shows statement true for \(n=1\) |
| Assume true for \(n=k\) so \(2^{3k+1}+3(5^{2k+1})\) is divisible by 17 | M1 | Assumes true for \(n=k\) |
| \(f(k+1)-f(k)=2^{3k+4}+3(5^{2k+3})-2^{3k+1}-3(5^{2k+1})\) | M1 | Attempts \(f(k+1)-f(k)\) |
| \(=7\times2^{3k+1}+7\times3(5^{2k+1})+17\times3(5^{2k+1})\) | ||
| \(=7f(k)+17\times3(5^{2k+1})\) | A1 | Correct expression in terms of \(f(k)\) |
| \(f(k+1)=8f(k)+17\times3(5^{2k+1})\) | A1 | Correct expression in terms of \(f(k)\) |
| If true for \(n=k\) then shown true for \(n=k+1\), and since true for \(n=1\), true for all positive integers \(n\) | A1 | Correct complete conclusion |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$, $2^{3n+1}+3(5^{2n+1})=16+375=391$; $391=17\times23$ so true for $n=1$ | B1 | Shows statement true for $n=1$ |
| Assume true for $n=k$ so $2^{3k+1}+3(5^{2k+1})$ is divisible by 17 | M1 | Assumes true for $n=k$ |
| $f(k+1)-f(k)=2^{3k+4}+3(5^{2k+3})-2^{3k+1}-3(5^{2k+1})$ | M1 | Attempts $f(k+1)-f(k)$ |
| $=7\times2^{3k+1}+7\times3(5^{2k+1})+17\times3(5^{2k+1})$ | | |
| $=7f(k)+17\times3(5^{2k+1})$ | A1 | Correct expression in terms of $f(k)$ |
| $f(k+1)=8f(k)+17\times3(5^{2k+1})$ | A1 | Correct expression in terms of $f(k)$ |
| If true for $n=k$ then shown true for $n=k+1$, and since true for $n=1$, true for all positive integers $n$ | A1 | Correct complete conclusion |
**(6 marks)**
---\begin{enumerate}
\item Prove by induction that for all positive integers $n$,
\end{enumerate}
$$f ( n ) = 2 ^ { 3 n + 1 } + 3 \left( 5 ^ { 2 n + 1 } \right)$$
is divisible by 17
\hfill \mbox{\textit{Edexcel CP1 Q2 [6]}}