| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core (Further Pure Core) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| 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 question requiring partial fractions decomposition, telescoping series manipulation, and limit evaluation. While the method of differences is a standard technique at this level, correctly setting up the telescoping sum, tracking terms through the cancellation, and determining the specific constants a and b requires careful algebraic manipulation beyond routine A-level work. The question is methodical rather than requiring deep insight, but the multi-step nature and Further Maths context place it moderately above average difficulty. |
| Spec | 4.05c Partial fractions: extended to quadratic denominators4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | 1 𝐴 𝐵 |
| Answer | Marks |
|---|---|
| 4 2(𝑛+1)(𝑛+2) | M1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1.1 | enough terms to show consistent cancellation in their series | |
| 3 | (b) | 3 |
| 4 | B1 | |
| [1] | 2.2a |
Question 3:
3 | (a) | 1 𝐴 𝐵
= +
𝑟(𝑟+2) 𝑟 (𝑟+2)
1 = 𝐴(𝑟+2)+𝐵𝑟
(cid:2869) (cid:2869)
𝑟 = 0 ⇒ 𝐴 = , 𝑟 = −2 ⇒ 𝐵 = −
(cid:2870) (cid:2870)
(cid:3041) (cid:3041)
1 1 1 1
(cid:3533) = (cid:3533)(cid:3436) − (cid:3440)
𝑟(𝑟+2) 2 𝑟 𝑟+2
(cid:3045)(cid:2880)(cid:2869) (cid:3045)(cid:2880)(cid:2869)
1 1 1 1 1 1 1
= 𝑘(cid:3428)1− + − + − +...+ − (cid:3432)
3 2 4 3 5 𝑛 𝑛+2
1 1 1 1
= (cid:3428)1+ − − (cid:3432)
2 2 𝑛+1 𝑛+2
3 2𝑛+3
= −
4 2(𝑛+1)(𝑛+2) | M1
A1
M1
A1
A1
[5] | 1.1
1.1
1.1
1.1
1.1 | enough terms to show consistent cancellation in their series
3 | (b) | 3
4 | B1
[1] | 2.2a3
\begin{enumerate}[label=(\alph*)]
\item Using partial fractions and the method of differences, show that
$$\frac { 1 } { 1 \times 3 } + \frac { 1 } { 2 \times 4 } + \frac { 1 } { 3 \times 5 } + \ldots + \frac { 1 } { \mathrm { n } ( \mathrm { n } + 2 ) } = \frac { 3 } { 4 } - \frac { \mathrm { an } + \mathrm { b } } { 2 ( \mathrm { n } + 1 ) ( \mathrm { n } + 2 ) }$$
where $a$ and $b$ are integers to be determined.
\item Deduce the sum to infinity of the series.
$$\frac { 1 } { 1 \times 3 } + \frac { 1 } { 2 \times 4 } + \frac { 1 } { 3 \times 5 } + \ldots$$
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q3 [6]}}