| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Convergence and Limits of Sequences |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on sequence convergence requiring induction proof, monotonicity analysis, and limit finding. While it involves multiple techniques (induction, algebraic manipulation, limits), each part is well-scaffolded and follows standard approaches. The induction and algebraic steps are routine for Further Maths students, making it moderately above average difficulty but not requiring exceptional insight. |
| Spec | 4.01a Mathematical induction: construct proofs8.01a Recurrence relations: general sequences, closed form and recurrence8.01c Sequence behaviour: periodic, convergent, divergent, oscillating, monotonic8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | u 2− 3 = −2 < 0 so result is true for n = 1 |
| Answer | Marks |
|---|---|
| the result for all positive integers n | B1 |
| Answer | Marks |
|---|---|
| [6] | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | Considering the (k + 1)th case |
| Answer | Marks |
|---|---|
| (b) | 2u +3 u (u +2) 3−u 2 |
| Answer | Marks |
|---|---|
| n + 1 n | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| (c) | 3−a2 |
| Answer | Marks |
|---|---|
| n + 1 n | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | Special Case B1 for correct limit with invalid or |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | Monotonic (or strictly) increasing, convergent | B1 |
| [1] | 2.5 |
Question 8:
8 | (a) | u 2− 3 = −2 < 0 so result is true for n = 1
1
Assume that u 2− 3 < 0 for some k ≥ 1
k
2
2u +3
Then u 2 −3= k −3
k+1 u +2
k
(2u +3)2 −3(u +2)2
= k k
(u +2)2
k
4u 2 +12u +9−3u 2 −12u −12 u 2 −3
= k k k k = k < 0
(u +2)2 (u +2)2
k k
Explanation that 1st case true and kth case true ⇒ (k + 1)th case true gives
the result for all positive integers n | B1
M1
M1
M1
A1
E1
[6] | 1.1
1.1
2.1
1.1
2.2a
2.4 | Considering the (k + 1)th case
Common denominator
Must be noted that it is negative
(b) | 2u +3 u (u +2) 3−u 2
u −u = n − n n = n
n + 1 n u +2 u +2 u +2
n n n
Explaining that u >u from (a)’s result
n + 1 n | M1
A1
B1
[3] | 3.1a
1.1
2.2a
(c) | 3−a2
=0⇒a =± 3
a+2
from (b) u >u so a = 3
n + 1 n | M1
A1
[2] | 3.1a
1.1 | Special Case B1 for correct limit with invalid or
missing reasoning
(d) | Monotonic (or strictly) increasing, convergent | B1
[1] | 2.5
PMT
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For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored8 The sequence $\left\{ u _ { n } \right\}$ of positive real numbers is defined by $u _ { 1 } = 1$ and $u _ { n + 1 } = \frac { 2 u _ { n } + 3 } { u _ { n } + 2 }$ for $n \geqslant 1$.
\begin{enumerate}[label=(\alph*)]
\item Prove by induction that $u _ { n } ^ { 2 } - 3 < 0$ for all positive integers $n$.
\item By considering $u _ { n + 1 } - u _ { n }$, use the result of part (a) to show that $u _ { n + 1 } > u _ { n }$ for all positive integers $n$.
The sequence $\left\{ u _ { n } \right\}$ has a limit for $n \rightarrow \infty$.
\item Find the limit of the sequence $\left\{ u _ { n } \right\}$ as $n \rightarrow \infty$.
\item Describe as fully as possible the behaviour of the sequence $\left\{ u _ { n } \right\}$.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2020 Q8 [12]}}