| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Partial Fractions and Telescoping Series |
| Difficulty | Challenging +1.2 This is a structured multi-part question requiring standard techniques: (a) uses given formulae for sum of squares with arithmetic progression, (b) applies routine partial fractions decomposition followed by telescoping series recognition, and (c) requires taking a limit. While it involves Further Maths content and multiple steps, each part follows predictable methods without requiring novel insight, placing it moderately above average difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_n = \sum_{r=1}^{n}(4r-2)^2 = 16\sum_{r=1}^{n}r^2 - 16\sum_{r=1}^{n}r + 4n\) | M1 A1 | |
| \(= \frac{8}{3}n(n+1)(2n+1) - 8n(n+1) + 4n\) | M1 | |
| \(\frac{4}{3}n(4n^2+6n+2-6n-6+3) = \frac{4}{3}n(4n^2-1)\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{n}{S_n} = \frac{3}{4(2n-1)(2n+1)} = \frac{3}{8}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)\) | M1 A1 | |
| \(\sum_{n=1}^{N}\frac{n}{S_n} = \frac{3}{8}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\ldots+\frac{1}{2N-1}-\frac{1}{2N+1}\right)\) | M1 | |
| \(= \frac{3}{8}\left(1-\frac{1}{2N+1}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{8}\) | B1 |
## Question 3:
**Part 3(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_n = \sum_{r=1}^{n}(4r-2)^2 = 16\sum_{r=1}^{n}r^2 - 16\sum_{r=1}^{n}r + 4n$ | M1 A1 | |
| $= \frac{8}{3}n(n+1)(2n+1) - 8n(n+1) + 4n$ | M1 | |
| $\frac{4}{3}n(4n^2+6n+2-6n-6+3) = \frac{4}{3}n(4n^2-1)$ | A1 | AG |
**Part 3(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{n}{S_n} = \frac{3}{4(2n-1)(2n+1)} = \frac{3}{8}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)$ | M1 A1 | |
| $\sum_{n=1}^{N}\frac{n}{S_n} = \frac{3}{8}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\ldots+\frac{1}{2N-1}-\frac{1}{2N+1}\right)$ | M1 | |
| $= \frac{3}{8}\left(1-\frac{1}{2N+1}\right)$ | A1 | |
**Part 3(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{8}$ | B1 | |
---3 Let $S _ { n } = 2 ^ { 2 } + 6 ^ { 2 } + 10 ^ { 2 } + \ldots + ( 4 n - 2 ) ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Use standard results from the List of Formulae (MF19) to show that $S _ { n } = \frac { 4 } { 3 } n \left( 4 n ^ { 2 } - 1 \right)$.
\item Express $\frac { \mathrm { n } } { \mathrm { S } _ { \mathrm { n } } }$ in partial fractions and find $\sum _ { \mathrm { n } = 1 } ^ { \mathrm { N } } \frac { \mathrm { n } } { \mathrm { S } _ { \mathrm { n } } }$ in terms of $N$.
\item Deduce the value of $\sum _ { n = 1 } ^ { \infty } \frac { n } { S _ { n } }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q3 [9]}}