| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove recurrence relation formula |
| Difficulty | Standard +0.8 This is a standard Further Maths induction proof with a recurrence relation, requiring calculation of terms and a multi-step inductive proof involving algebraic manipulation. While routine for FP1, it's above average difficulty due to the algebraic complexity in the inductive step and being a Further Maths topic. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(u_2 = \frac{2}{1+2} = \frac{2}{3},\ u_3 = \frac{\frac{2}{3}}{1+\frac{2}{3}} = \frac{2}{5}\) | M1, A1 [2] | Use of inductive definition; c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(n=1\), \(\frac{2}{2\times1-1} = 2\), so true for \(n=1\) | B1 | Showing use of \(u_n = \frac{2}{2n-1}\) |
| Assume \(u_k = \frac{2}{2k-1}\) | E1 | Assuming true for \(k\) |
| \(\Rightarrow u_{k+1} = \frac{\frac{2}{2k-1}}{1+\frac{2}{2k-1}}\) | M1 | \(u_{k+1}\) |
| \(= \frac{\frac{2}{2k-1}}{\frac{2k-1+2}{2k-1}} = \frac{2}{2k+1}\) | A1 | Correct simplification |
| \(= \frac{2}{2(k+1)-1}\) | ||
| This is the given result with \(k+1\) replacing \(k\). Therefore if true for \(k\) it is also true for \(k+1\). Since true for \(k=1\), true for all positive integers. | E1, E1 [6] | Dependent on A1 and previous E1; dependent on B1 and previous E1 |
# Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $u_2 = \frac{2}{1+2} = \frac{2}{3},\ u_3 = \frac{\frac{2}{3}}{1+\frac{2}{3}} = \frac{2}{5}$ | M1, A1 **[2]** | Use of inductive definition; c.a.o. |
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# Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| When $n=1$, $\frac{2}{2\times1-1} = 2$, so true for $n=1$ | B1 | Showing use of $u_n = \frac{2}{2n-1}$ |
| Assume $u_k = \frac{2}{2k-1}$ | E1 | Assuming true for $k$ |
| $\Rightarrow u_{k+1} = \frac{\frac{2}{2k-1}}{1+\frac{2}{2k-1}}$ | M1 | $u_{k+1}$ |
| $= \frac{\frac{2}{2k-1}}{\frac{2k-1+2}{2k-1}} = \frac{2}{2k+1}$ | A1 | Correct simplification |
| $= \frac{2}{2(k+1)-1}$ | | |
| This is the given result with $k+1$ replacing $k$. Therefore if true for $k$ it is also true for $k+1$. Since true for $k=1$, true for all positive integers. | E1, E1 **[6]** | Dependent on A1 and previous E1; dependent on B1 and previous E1 |
---6 A sequence is defined by $u _ { 1 } = 2$ and $u _ { n + 1 } = \frac { u _ { n } } { 1 + u _ { n } }$.\\
(i) Calculate $u _ { 3 }$.\\
(ii) Prove by induction that $u _ { n } = \frac { 2 } { 2 n - 1 }$.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI FP1 2010 Q6 [8]}}