| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove recurrence relation formula |
| Difficulty | Standard +0.3 This is a straightforward induction proof with a given formula. Part (i) requires simple recursive calculation (a₂=6, a₃=21). Part (ii) follows standard induction structure: base case is trivial verification, and the inductive step involves substituting the recurrence relation into the assumed formula—algebraically routine with no conceptual obstacles. Slightly above average difficulty only because it's Further Maths and requires formal proof writing, but the mechanics are entirely standard. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| \(a_2 = 3 \times 2 = 6,\ a_3 = 3 \times 7 = 21\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| When \(n=1\), \(\frac{5 \times 3^0 - 3}{2} = 1\), so true for \(n=1\) | B1 | Showing use of \(a_n = \frac{5 \times 3^{n-1}-3}{2}\) |
| Assume \(a_k = \frac{5 \times 3^{k-1}-3}{2}\) | E1 | Assuming true for \(n = k\) |
| \(\Rightarrow a_{k+1} = 3\left(\frac{5 \times 3^{k-1}-3}{2}+1\right)\) | M1 | \(a_{k+1}\), using \(a_k\) and attempting to simplify |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{5 \times 3^k - 3}{2} = \frac{5 \times 3^{(k+1)-1}-3}{2}\) | A1 | Correct simplification to left hand expression |
| But this is the given result with \(k+1\) replacing \(k\). Therefore if true for \(n=k\) it is also true for \(n=k+1\). Since true for \(n=1\), true for all positive integers. | E1, E1 | Dependent on A1 and previous E1; Dependent on B1 and previous E1 |
# Question 6:
## Part (i):
$a_2 = 3 \times 2 = 6,\ a_3 = 3 \times 7 = 21$ | B1 | cao
**[1]**
## Part (ii):
When $n=1$, $\frac{5 \times 3^0 - 3}{2} = 1$, so true for $n=1$ | B1 | Showing use of $a_n = \frac{5 \times 3^{n-1}-3}{2}$
Assume $a_k = \frac{5 \times 3^{k-1}-3}{2}$ | E1 | Assuming true for $n = k$
$\Rightarrow a_{k+1} = 3\left(\frac{5 \times 3^{k-1}-3}{2}+1\right)$ | M1 | $a_{k+1}$, using $a_k$ and attempting to simplify
$= \frac{5 \times 3^k - 9}{2} + 3 = \frac{5 \times 3^k - 9 + 6}{2}$
$= \frac{5 \times 3^k - 3}{2} = \frac{5 \times 3^{(k+1)-1}-3}{2}$ | A1 | Correct simplification to left hand expression
But this is the given result with $k+1$ replacing $k$. Therefore if true for $n=k$ it is also true for $n=k+1$. Since true for $n=1$, true for all positive integers. | E1, E1 | Dependent on A1 and previous E1; Dependent on B1 and previous E1
**[6]**
---6 A sequence is defined by $a _ { 1 } = 1$ and $a _ { k + 1 } = 3 \left( a _ { k } + 1 \right)$.\\
(i) Calculate the value of the third term, $a _ { 3 }$.\\
(ii) Prove by induction that $a _ { n } = \frac { 5 \times 3 ^ { n - 1 } - 3 } { 2 }$.
\hfill \mbox{\textit{OCR MEI FP1 2012 Q6 [7]}}