| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2020 |
| 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 requires proper inductive structure, the algebraic steps are mechanical and this is a textbook-style divisibility proof, making it slightly easier than average. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(n=1\), \(2^{n+2} + 3^{2n+1} = 2^3 + 3^3 = 35\); shows statement true for \(n=1\) | B1 | Shows \(f(1)=35\) and concludes divisible by 7 |
| Assume true for \(n=k\), so \(2^{k+2} + 3^{2k+1}\) is divisible by 7 | M1 | Makes assumption statement for some value of \(n\) |
| \(f(k+1) - f(k) = 2^{k+3} + 3^{2k+3} - (2^{k+2} + 3^{2k+1})\) | M1 | Attempts \(f(k+1) - f(k)\) |
| \(= 2\times2^{k+2} + 9\times3^{2k+1} - 2^{k+2} - 3^{2k+1}\) \(= 2^{k+2} + 8\times3^{2k+1}\) \(= f(k) + 7\times3^{2k+1}\) or \(8f(k) - 7\times2^{k+2}\) | A1 | Achieves correct expression for \(f(k+1)-f(k)\) in terms of \(f(k)\) |
| \(f(k+1) = 2f(k) + 7\times3^{2k+1}\) or \(9f(k) - 7\times2^{k+2}\) | A1 | Reaches correct expression for \(f(k+1)\) in terms of \(f(k)\) |
| If true for \(n=k\) then true for \(n=k+1\) and as true for \(n=1\), true for all positive integers \(n\) | A1 | Correct conclusion, dependent on all previous marks; must convey all four bold ideas |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(n=1\), \(2^{n+2}+3^{2n+1}=2^3+3^3=35\); true for \(n=1\) | B1 | Shows \(f(1)=35\) and concludes divisible by 7 |
| Assume true for \(n=k\), so \(2^{k+2}+3^{2k+1}\) is divisible by 7 | M1 | Makes assumption for some value of \(n\) |
| \(f(k+1) = 2^{(k+1)+2} + 3^{2(k+1)+1}\) | M1 | Attempts \(f(k+1)\) |
| \(f(k+1) = 2^{k+3}+3^{2k+3} = 2\times2^{k+2}+9\times3^{2k+1}\) \(= 2(2^{k+2}+3^{2k+1})+7\times3^{2k+1}\) \(= 2f(k)+7\times3^{2k+1}\) or \(9f(k)-7\times2^{k+2}\) | A1, A1 | Correctly obtains either \(2f(k)\) or \(7\times3^{2k+1}\); reaches correct expression for \(f(k+1)\) in terms of \(f(k)\) |
| If true for \(n=k\) then true for \(n=k+1\) and as true for \(n=1\), true for all positive integers \(n\) | A1 | Correct conclusion, dependent on all previous marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(n=1\), \(2^{n+2}+3^{2n+1}=35\); true for \(n=1\) | B1 | Shows \(f(1)=35\) and concludes divisible by 7 |
| Assume true for \(n=k\), so \(2^{k+2}+3^{2k+1}\) is divisible by 7 | M1 | Makes assumption for some value of \(n\) |
| \(f(k+1)-mf(k) = 2^{k+3}+3^{2k+3}-m(2^{k+2}+3^{2k+1})\) | M1 | Attempts \(f(k+1)-mf(k)\) |
| \(= 2\times2^{k+2}+9\times3^{2k+1}-m\times2^{k+2}-m\times3^{2k+1}\) \(= (2-m)2^{k+2}+9\times3^{2k+1}-m\times3^{2k+1}\) \(= (2-m)(2^{k+2}+3^{2k+1})+7\times3^{2k+1}\) | A1 | Achieves correct expression for \(f(k+1)-mf(k)\) in terms of \(f(k)\) |
| \(f(k+1) = (2-m)(2^{k+2}+3^{2k+1})+7\times3^{2k+1}+mf(k)\) | A1 | Reaches correct expression for \(f(k+1)\) in terms of \(f(k)\) |
| If true for \(n=k\) then true for \(n=k+1\) and as true for \(n=1\), true for all positive integers \(n\) | A1 | Correct conclusion, dependent on all previous marks |
## Question 8:
**Way 1: $f(k+1) - f(k)$**
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$, $2^{n+2} + 3^{2n+1} = 2^3 + 3^3 = 35$; shows statement true for $n=1$ | B1 | Shows $f(1)=35$ and concludes divisible by 7 |
| Assume true for $n=k$, so $2^{k+2} + 3^{2k+1}$ is divisible by 7 | M1 | Makes assumption statement for some value of $n$ |
| $f(k+1) - f(k) = 2^{k+3} + 3^{2k+3} - (2^{k+2} + 3^{2k+1})$ | M1 | Attempts $f(k+1) - f(k)$ |
| $= 2\times2^{k+2} + 9\times3^{2k+1} - 2^{k+2} - 3^{2k+1}$ $= 2^{k+2} + 8\times3^{2k+1}$ $= f(k) + 7\times3^{2k+1}$ or $8f(k) - 7\times2^{k+2}$ | A1 | Achieves correct expression for $f(k+1)-f(k)$ in terms of $f(k)$ |
| $f(k+1) = 2f(k) + 7\times3^{2k+1}$ or $9f(k) - 7\times2^{k+2}$ | A1 | Reaches correct expression for $f(k+1)$ in terms of $f(k)$ |
| If true for $n=k$ then true for $n=k+1$ and as true for $n=1$, true for all positive integers $n$ | A1 | Correct conclusion, dependent on all previous marks; must convey all four bold ideas |
**Way 2: $f(k+1)$**
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$, $2^{n+2}+3^{2n+1}=2^3+3^3=35$; true for $n=1$ | B1 | Shows $f(1)=35$ and concludes divisible by 7 |
| Assume true for $n=k$, so $2^{k+2}+3^{2k+1}$ is divisible by 7 | M1 | Makes assumption for some value of $n$ |
| $f(k+1) = 2^{(k+1)+2} + 3^{2(k+1)+1}$ | M1 | Attempts $f(k+1)$ |
| $f(k+1) = 2^{k+3}+3^{2k+3} = 2\times2^{k+2}+9\times3^{2k+1}$ $= 2(2^{k+2}+3^{2k+1})+7\times3^{2k+1}$ $= 2f(k)+7\times3^{2k+1}$ or $9f(k)-7\times2^{k+2}$ | A1, A1 | Correctly obtains either $2f(k)$ or $7\times3^{2k+1}$; reaches correct expression for $f(k+1)$ in terms of $f(k)$ |
| If true for $n=k$ then true for $n=k+1$ and as true for $n=1$, true for all positive integers $n$ | A1 | Correct conclusion, dependent on all previous marks |
**Way 3: $f(k+1) - mf(k)$**
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$, $2^{n+2}+3^{2n+1}=35$; true for $n=1$ | B1 | Shows $f(1)=35$ and concludes divisible by 7 |
| Assume true for $n=k$, so $2^{k+2}+3^{2k+1}$ is divisible by 7 | M1 | Makes assumption for some value of $n$ |
| $f(k+1)-mf(k) = 2^{k+3}+3^{2k+3}-m(2^{k+2}+3^{2k+1})$ | M1 | Attempts $f(k+1)-mf(k)$ |
| $= 2\times2^{k+2}+9\times3^{2k+1}-m\times2^{k+2}-m\times3^{2k+1}$ $= (2-m)2^{k+2}+9\times3^{2k+1}-m\times3^{2k+1}$ $= (2-m)(2^{k+2}+3^{2k+1})+7\times3^{2k+1}$ | A1 | Achieves correct expression for $f(k+1)-mf(k)$ in terms of $f(k)$ |
| $f(k+1) = (2-m)(2^{k+2}+3^{2k+1})+7\times3^{2k+1}+mf(k)$ | A1 | Reaches correct expression for $f(k+1)$ in terms of $f(k)$ |
| If true for $n=k$ then true for $n=k+1$ and as true for $n=1$, true for all positive integers $n$ | A1 | Correct conclusion, dependent on all previous marks |
---\begin{enumerate}
\item Prove by induction that, for $n \in \mathbb { Z } ^ { + }$
\end{enumerate}
$$f ( n ) = 2 ^ { n + 2 } + 3 ^ { 2 n + 1 }$$
is divisible by 7
\hfill \mbox{\textit{Edexcel CP AS 2020 Q8 [6]}}