| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Standard +0.8 This is a Further Maths FP1 question requiring partial fractions decomposition, telescoping series recognition, and infinite series convergence. While the algebraic verification in (i) is routine, parts (ii) and (iii) demand understanding of telescoping sums and limit behavior—skills beyond standard A-level. The multi-step nature and conceptual depth place it moderately above average difficulty. |
| Spec | 1.04j Sum to infinity: convergent geometric series |r|<14.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\text{LHS} = \frac{2r+1-(2r-1)}{(2r-1)(2r+1)} = \frac{2}{4r^2-1} = \text{RHS}\) | M1, A1 | For correct process for adding fractions / For showing given result correctly |
| Answer | Marks | Guidance |
|---|---|---|
| This is \(1 - \frac{1}{2n+1}\) | M1, A1, M1, A1 | For expressing terms as differences using (i) / For at least first two and last terms correct / For cancelling pairs of terms / For any correct form |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) (a) Sum to infinity is 1 | B1 | For correct value; follow their (ii) if cvgyt |
| (b) Required sum is \(\frac{1}{2n+1}\) | B1 | For correct difference of their (iii)(a) and (ii) |
**(i)** $\text{LHS} = \frac{2r+1-(2r-1)}{(2r-1)(2r+1)} = \frac{2}{4r^2-1} = \text{RHS}$ | M1, A1 | For correct process for adding fractions / For showing given result correctly
**Total: 2 marks**
**(ii)** Sum is $\left(1-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{5}\right) + \left(\frac{1}{5}-\frac{1}{7}\right) + \cdots + \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$
This is $1 - \frac{1}{2n+1}$ | M1, A1, M1, A1 | For expressing terms as differences using (i) / For at least first two and last terms correct / For cancelling pairs of terms / For any correct form
**Total: 4 marks**
**(iii)** **(a)** Sum to infinity is 1 | B1 | For correct value; follow their (ii) if cvgyt
**(b)** Required sum is $\frac{1}{2n+1}$ | B1 | For correct difference of their (iii)(a) and (ii)
**Total: 2 marks**
**Question 5 Total: 8 marks**
---5 (i) Show that
$$\frac { 1 } { 2 r - 1 } - \frac { 1 } { 2 r + 1 } = \frac { 2 } { 4 r ^ { 2 } - 1 }$$
(ii) Hence find an expression in terms of $n$ for
$$\frac { 2 } { 3 } + \frac { 2 } { 15 } + \frac { 2 } { 35 } + \ldots + \frac { 2 } { 4 n ^ { 2 } - 1 }$$
(iii) State the value of
\begin{enumerate}[label=(\alph*)]
\item $\quad \sum _ { r = 1 } ^ { \infty } \frac { 2 } { 4 r ^ { 2 } - 1 }$,
\item $\quad \sum _ { r = n + 1 } ^ { \infty } \frac { 2 } { 4 r ^ { 2 } - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR FP1 Q5 [8]}}