| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove sequence property via recurrence |
| Difficulty | Standard +0.3 This is a straightforward induction proof with a helpful recurrence relation provided. Part (i) involves routine algebraic manipulation of exponentials (expanding f(n+1) and simplifying), while part (ii) is a standard divisibility induction where the recurrence relation makes the inductive step trivial since 28 and 7 are clearly divisible by 7. The question requires no novel insight—just following the standard induction template with algebraic care. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(n)+f(n+1) = 3^{3n}+6^{n-1}+3^{3n+3}+6^n\) | M1 | Establishes initial result |
| \(= 3^{3n}(1+27)+6^{n-1}(1+6)\) | A1 | |
| \(= 28(3^{3n})+7(6^{n-1})\) (AG) | Part mark: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_k\): \(f(k) = 7\lambda\) | B1 | States inductive hypothesis |
| \(3^3 + 6^0 = 28 = 4\times7 \Rightarrow H_1\) is true | B1 | Proves base case |
| \(f(k+1)+f(k) = f(k+1)+7\lambda = 28(3^{3k})+7(6^{k-1}) = 7\mu\) | M1 | Shows \(P_k \Rightarrow P_{k+1}\) |
| \(\Rightarrow f(k+1) = 7(\mu-\lambda) \therefore H_k \Rightarrow H_{k+1}\) | ||
| Hence by the principle of mathematical induction \(H_n\) is true for all positive integers \(n\) | A1 | States conclusion; Part mark: 4 |
## Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(n)+f(n+1) = 3^{3n}+6^{n-1}+3^{3n+3}+6^n$ | M1 | Establishes initial result |
| $= 3^{3n}(1+27)+6^{n-1}(1+6)$ | A1 | |
| $= 28(3^{3n})+7(6^{n-1})$ (AG) | | Part mark: 2 |
## Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_k$: $f(k) = 7\lambda$ | B1 | States inductive hypothesis |
| $3^3 + 6^0 = 28 = 4\times7 \Rightarrow H_1$ is true | B1 | Proves base case |
| $f(k+1)+f(k) = f(k+1)+7\lambda = 28(3^{3k})+7(6^{k-1}) = 7\mu$ | M1 | Shows $P_k \Rightarrow P_{k+1}$ |
| $\Rightarrow f(k+1) = 7(\mu-\lambda) \therefore H_k \Rightarrow H_{k+1}$ | | |
| Hence by the principle of mathematical induction $H_n$ is **true for all positive integers** $n$ | A1 | States conclusion; Part mark: 4 |
---4 It is given that $\mathrm { f } ( n ) = 3 ^ { 3 n } + 6 ^ { n - 1 }$.\\
(i) Show that $\mathrm { f } ( n + 1 ) + \mathrm { f } ( n ) = 28 \left( 3 ^ { 3 n } \right) + 7 \left( 6 ^ { n - 1 } \right)$.\\
(ii) Hence, or otherwise, prove by mathematical induction that $\mathrm { f } ( n )$ is divisible by 7 for every positive integer $n$.
\hfill \mbox{\textit{CAIE FP1 2011 Q4 [6]}}