| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on telescoping series where the partial fraction decomposition is given. Part (i) requires applying the method of differences (a standard FP1 technique), part (ii) is straightforward limit evaluation, and part (iii) requires identifying the correct range of r values. While it involves multiple steps and careful algebraic manipulation, the method is prescribed and follows a well-practiced template for FP1 students. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks |
|---|---|
| Write out telescoping sum using partial fractions identity | M1 M1 |
| Most terms cancel; remaining terms give \(\frac{2}{3} - \frac{3}{4(2n+1)}+\frac{1}{4(2n+3)}\) | A1 A1 A1 A1 |
| Answer | Marks |
|---|---|
| As \(n\to\infty\), limit \(= \frac{2}{3}\) | B1 |
| Answer | Marks |
|---|---|
| Identify first term: \(r=20\) (since \(2r-1=39 \Rightarrow r=20\)), last term \(r=50\) | M1 M1 |
| \(S = \left[\frac{2}{3}-\frac{3}{4(101)}+\frac{1}{4(103)}\right] - \left[\frac{2}{3}-\frac{3}{4(39)}+\frac{1}{4(41)}\right]\) | M1 |
| \(= \frac{3}{156}-\frac{3}{404}+\frac{1}{4(41)}-\frac{1}{412}\) evaluated \(\approx 0.0129\) | A1 |
## Question 9:
**(i)**
Write out telescoping sum using partial fractions identity | M1 M1 |
Most terms cancel; remaining terms give $\frac{2}{3} - \frac{3}{4(2n+1)}+\frac{1}{4(2n+3)}$ | A1 A1 A1 A1 |
**(ii)**
As $n\to\infty$, limit $= \frac{2}{3}$ | B1 |
**(iii)**
Identify first term: $r=20$ (since $2r-1=39 \Rightarrow r=20$), last term $r=50$ | M1 M1 |
$S = \left[\frac{2}{3}-\frac{3}{4(101)}+\frac{1}{4(103)}\right] - \left[\frac{2}{3}-\frac{3}{4(39)}+\frac{1}{4(41)}\right]$ | M1 |
$= \frac{3}{156}-\frac{3}{404}+\frac{1}{4(41)}-\frac{1}{412}$ evaluated $\approx 0.0129$ | A1 |9 You are given that $\frac { 3 } { 4 ( 2 r - 1 ) } - \frac { 1 } { 2 r + 1 } + \frac { 1 } { 4 ( 2 r + 3 ) } = \frac { 2 r + 5 } { ( 2 r - 1 ) ( 2 r + 1 ) ( 2 r + 3 ) }$.\\
(i) Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } \frac { 2 r + 5 } { ( 2 r - 1 ) ( 2 r + 1 ) ( 2 r + 3 ) } = \frac { 2 } { 3 } - \frac { 3 } { 4 ( 2 n + 1 ) } + \frac { 1 } { 4 ( 2 n + 3 ) } .$$
(ii) Write down the limit to which $\sum _ { r = 1 } ^ { n } \frac { 2 r + 5 } { ( 2 r - 1 ) ( 2 r + 1 ) ( 2 r + 3 ) }$ converges as $n$ tends to infinity.\\
(iii) Find the sum of the finite series
$$\frac { 45 } { 39 \times 41 \times 43 } + \frac { 47 } { 41 \times 43 \times 45 } + \frac { 49 } { 43 \times 45 \times 47 } + \ldots + \frac { 105 } { 99 \times 101 \times 103 } ,$$
giving your answer to 3 significant figures.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR MEI FP1 2016 Q9 [11]}}