| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 8 |
| 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) is trivial calculation. Part (ii) requires standard induction steps: verify base case, assume for n=k, prove for n=k+1 using the recurrence relation. The algebra is routine and the formula is provided, making this slightly easier than average for Further Maths induction questions. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a_2 = 7\times7 - 3 = 46\) | M1 | Use of inductive definition |
| \(a_3 = 7\times46 - 3 = 319\) | A1 [2] | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(n=1\), \(\frac{13\times7^0+1}{2} = 7\), so true for \(n=1\) | B1 | Correct use of part (i) (may be implied) |
| Assume true for \(n=k\): \(a_k = \frac{13\times7^{k-1}+1}{2}\) | E1 | Assuming true for \(k\) |
| \(\Rightarrow a_{k+1} = 7\times\frac{13\times7^{k-1}+1}{2} - 3\) | M1 | Attempt to use \(a_{k+1} = 7a_k - 3\) |
| \(= \frac{13\times7^k + 7}{2} - 3\) | ||
| \(= \frac{13\times7^k + 7 - 6}{2}\) | ||
| \(= \frac{13\times7^k + 1}{2}\) | A1 | Correct simplification |
| But this is the given result with \(k+1\) replacing \(k\). Therefore if true for \(k\) it is true for \(k+1\). Since true for \(k=1\), true for \(k=1,2,3\) and so all positive integers. | E1 | Dependent on A1 and previous E1 |
| E1 [6] | Dependent on B1 and previous E1 |
# Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a_2 = 7\times7 - 3 = 46$ | M1 | Use of inductive definition |
| $a_3 = 7\times46 - 3 = 319$ | A1 **[2]** | c.a.o. |
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# Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$, $\frac{13\times7^0+1}{2} = 7$, so true for $n=1$ | B1 | Correct use of part (i) (may be implied) |
| Assume true for $n=k$: $a_k = \frac{13\times7^{k-1}+1}{2}$ | E1 | Assuming true for $k$ |
| $\Rightarrow a_{k+1} = 7\times\frac{13\times7^{k-1}+1}{2} - 3$ | M1 | Attempt to use $a_{k+1} = 7a_k - 3$ |
| $= \frac{13\times7^k + 7}{2} - 3$ | | |
| $= \frac{13\times7^k + 7 - 6}{2}$ | | |
| $= \frac{13\times7^k + 1}{2}$ | A1 | Correct simplification |
| But this is the given result with $k+1$ replacing $k$. Therefore if true for $k$ it is true for $k+1$. Since true for $k=1$, true for $k=1,2,3$ and so all positive integers. | E1 | Dependent on A1 and previous E1 |
| | E1 **[6]** | Dependent on B1 and previous E1 |
---6 A sequence is defined by $a _ { 1 } = 7$ and $a _ { k + 1 } = 7 a _ { k } - 3$.\\
(i) Calculate the value of the third term, $a _ { 3 }$.\\
(ii) Prove by induction that $a _ { n } = \frac { \left( 13 \times 7 ^ { n - 1 } \right) + 1 } { 2 }$.
\hfill \mbox{\textit{OCR MEI FP1 2008 Q6 [8]}}