| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Challenging +1.2 This is a telescoping series question requiring recognition of the pattern and careful algebraic manipulation. While the expressions look complex with nested radicals, part (i) is computational (calculator work), and part (ii) requires finding when the partial sum exceeds a threshold. The key insight—that consecutive terms cancel—is a standard Further Maths technique, making this moderately above average but not requiring deep conceptual innovation. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks |
|---|---|
| (ii) | 6 7 7 8 35 36 36 |
| Answer | Marks |
|---|---|
| ⇒n>54.42K so 55 terms required. | M1A1 |
Total
| Answer | Marks | Guidance |
|---|---|---|
| Page 5 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2015 | 9231 | 11 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Solution | Marks |
Question 4:
--- 4 (i)
(ii) ---
4 (i)
(ii) | 6 7 7 8 35 36 36
− + − +K+ − =6− =4.820
1 3 3 7 871 931 931
n+6
6− >4.9⇒0.21n2−10.79n−34.79(>0)
n2+n+1
⇒n>54.42K so 55 terms required. | M1A1
A1
[3]
M1A1
dM1A1
[4]
Total
7
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9231 | 11
Qn &
Part | Solution | MarksThe sequence $a_1, a_2, a_3, \ldots$ is such that, for all positive integers $n$,
$$a_n = \frac{n + 5}{\sqrt{(n^2 - n + 1)}} - \frac{n + 6}{\sqrt{(n^2 + n + 1)}}.$$
The sum $\sum_{n=1}^{N} a_n$ is denoted by $S_N$. Find
\begin{enumerate}[label=(\roman*)]
\item the value of $S_{30}$ correct to 3 decimal places, [3]
\item the least value of $N$ for which $S_N > 4.9$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP1 2015 Q4 [7]}}