| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | September |
| Marks | 5 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Standard +0.8 This is a standard telescoping series question requiring partial fractions decomposition and recognition of cancellation patterns. While it involves infinite series convergence (a Further Maths topic), the technique is well-established and the algebraic manipulation is straightforward. The two-part structure guides students through the finite sum before the limit, making it accessible but still requiring solid understanding of series convergence. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{1}{(2r-1)(2r+1)} = \frac{A}{(2r-1)} + \frac{B}{(2r+1)} = \frac{1}{2}(\frac{1}{(2r-1)} - \frac{1}{(2r+1)})\) | M1, A1 | Partial fractions |
| \(= \frac{1}{2}(\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \ldots + \frac{1}{2n-1} - \frac{1}{2n+1})\) | M1 | Use method of differences |
| \(= \frac{1}{2}(1 - \frac{1}{2n+1}) = \frac{n}{2n+1}\) | A1 [4] | |
| (ii) e.g. \(\frac{n}{2n+1} = \frac{1}{2} \times \frac{2n+1-1}{2n+1} = \frac{1}{2}(1 - \frac{1}{2n+1}) \to \frac{1}{2}\) as \(n \to \infty\) | B1ft, [1] | FT from their answer providing it is of the form \(\frac{an+b}{cn+d}\) |
**(i)** $\frac{1}{(2r-1)(2r+1)} = \frac{A}{(2r-1)} + \frac{B}{(2r+1)} = \frac{1}{2}(\frac{1}{(2r-1)} - \frac{1}{(2r+1)})$ | M1, A1 | Partial fractions
$= \frac{1}{2}(\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \ldots + \frac{1}{2n-1} - \frac{1}{2n+1})$ | M1 | Use method of differences
$= \frac{1}{2}(1 - \frac{1}{2n+1}) = \frac{n}{2n+1}$ | A1 [4]
**(ii)** e.g. $\frac{n}{2n+1} = \frac{1}{2} \times \frac{2n+1-1}{2n+1} = \frac{1}{2}(1 - \frac{1}{2n+1}) \to \frac{1}{2}$ as $n \to \infty$ | B1ft, [1] | FT from their answer providing it is of the form $\frac{an+b}{cn+d}$
---6 (i) Find as a single algebraic fraction an expression for $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) }$.\\
(ii) Determine the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) }$.
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q6 [5]}}