| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Simple recurrence evaluation |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths question requiring standard limit evaluation (n/(n+1) ā 1 as nāā) and a simple monotonicity proof (showing a_{n+1} > a_n or using differences/ratios). Both parts are direct applications of basic techniques with no conceptual challenges, making it slightly easier than average despite being Further Maths content. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1\) | B1 | Cao. Must be clear that the limit is equal to 1. Tending to or approaching 1 or 0.999 or \(\to 1\) etc do not score |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a_n = 1 - \frac{1}{n+1}\) | M1 | |
| \(n+1\) increases as \(n\) increases so \(\frac{1}{n+1}\) decreases | M1 | Convincing explanation that \(\frac{1}{n+1}\) decreases |
| Less is being taken away from 1 each time, so \(a_n = 1 - \frac{1}{n+1}\) increases | E1 | Convincing completion (A.G.) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{n+1}{n+2}\) | M1 | Testing \(\frac{n}{n+1} < \frac{n+1}{n+2}\); formulating expression for \((n+1)\)th or \((n-1)\)th term |
| \(n(n+2) < (n+1)^2\) | M1 | Formulating inequality and cross multiplying |
| Expanding: \(n(n+2) = n^2+2n\) and \((n+1)^2 = n^2+2n+1\), convincing completion | E1 | Probably see \(\frac{n+1}{n+2} > \frac{n}{n+1}\) [i.e. increasing] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{n+1}{n+2}\) | M1 | Formulating expression for \((n+1)\)th term |
| Difference is \(\frac{n+1}{n+2} - \frac{n}{n+1} = \frac{(n+1)^2 - n(n+2)}{(n+1)(n+2)}\) | M1 | Finding difference and attempting to form single fraction |
| \(\frac{(n+1)^2 - n(n+2)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)} > 0\), so sequence is increasing | E1 | Convincing completion (A.G.) |
# Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1$ | B1 | Cao. Must be clear that the limit is **equal to 1**. Tending to or approaching 1 or 0.999 or $\to 1$ etc do not score |
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# Question 3(b):
**Main method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $a_n = 1 - \frac{1}{n+1}$ | M1 | |
| $n+1$ increases as $n$ increases so $\frac{1}{n+1}$ decreases | M1 | Convincing explanation that $\frac{1}{n+1}$ decreases |
| Less is being taken away from 1 each time, so $a_n = 1 - \frac{1}{n+1}$ increases | E1 | Convincing completion (A.G.) |
**Special Case** (Max 2): Correct differentiation and states $> 0$ to demonstrate increasing **SC B1**; goes on to consider positive integers as a subset of reals **SC B1** dep on first B1.
**Alternative method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{n+1}{n+2}$ | M1 | Testing $\frac{n}{n+1} < \frac{n+1}{n+2}$; formulating expression for $(n+1)$th or $(n-1)$th term |
| $n(n+2) < (n+1)^2$ | M1 | Formulating inequality and cross multiplying |
| Expanding: $n(n+2) = n^2+2n$ and $(n+1)^2 = n^2+2n+1$, convincing completion | E1 | Probably see $\frac{n+1}{n+2} > \frac{n}{n+1}$ [i.e. increasing] |
**Alternative method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{n+1}{n+2}$ | M1 | Formulating expression for $(n+1)$th term |
| Difference is $\frac{n+1}{n+2} - \frac{n}{n+1} = \frac{(n+1)^2 - n(n+2)}{(n+1)(n+2)}$ | M1 | Finding difference and attempting to form single fraction |
| $\frac{(n+1)^2 - n(n+2)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)} > 0$, so sequence is increasing | E1 | Convincing completion (A.G.) |
---3 An infinite sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is defined by $a _ { \mathrm { n } } = \frac { \mathrm { n } } { \mathrm { n } + 1 }$, for all positive integers $n$.
\begin{enumerate}[label=(\alph*)]
\item Find the limit of the sequence.
\item Prove that this is an increasing sequence.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2022 Q3 [4]}}