| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Finding constants from given sum formula |
| Difficulty | Standard +0.3 This is a standard Further Maths question on series summation using familiar techniques. Part (a) requires expanding and applying standard formulae for Σr and Σr², part (b) uses method of differences (a core FM technique), and part (c) is a straightforward limit. All three parts follow well-established procedures with no novel insight required, making it slightly easier than average for Further Maths content. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{n}(7r+1)(7r+8) = \sum_{r=1}^{n} 49r^2 + 63r + 8\) | M1 | Expands |
| \(= 49\left(\frac{1}{6}n(n+1)(2n+1)\right) + 63\left(\frac{1}{2}n(n+1)\right) + 8n\) | M1 | Substitutes formulae for \(\sum r^2\) and \(\sum r\) |
| \(= \frac{49}{3}n^3 + 56n^2 + \frac{143}{3}n\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{(7r+1)(7r+8)} = \frac{1}{7}\left(\frac{1}{7r+1} - \frac{1}{7r+8}\right)\) | M1 A1 | Finds partial fractions |
| \(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)} = \frac{1}{7}\left(\frac{1}{8} - \frac{1}{15} + \frac{1}{15} - \frac{1}{22} + \cdots + \frac{1}{7n+1} - \frac{1}{7n+8}\right)\) | M1 | Writes at least three correct terms, including first and last |
| \(= \frac{1}{7}\left(\frac{1}{8} - \frac{1}{7n+8}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{56}\) | B1 FT |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n}(7r+1)(7r+8) = \sum_{r=1}^{n} 49r^2 + 63r + 8$ | M1 | Expands |
| $= 49\left(\frac{1}{6}n(n+1)(2n+1)\right) + 63\left(\frac{1}{2}n(n+1)\right) + 8n$ | M1 | Substitutes formulae for $\sum r^2$ and $\sum r$ |
| $= \frac{49}{3}n^3 + 56n^2 + \frac{143}{3}n$ | A1 | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{(7r+1)(7r+8)} = \frac{1}{7}\left(\frac{1}{7r+1} - \frac{1}{7r+8}\right)$ | M1 A1 | Finds partial fractions |
| $\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)} = \frac{1}{7}\left(\frac{1}{8} - \frac{1}{15} + \frac{1}{15} - \frac{1}{22} + \cdots + \frac{1}{7n+1} - \frac{1}{7n+8}\right)$ | M1 | Writes at least three correct terms, including first and last |
| $= \frac{1}{7}\left(\frac{1}{8} - \frac{1}{7n+8}\right)$ | A1 | |
## Question 2(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{56}$ | B1 FT | |
---2
\begin{enumerate}[label=(\alph*)]
\item Use standard results from the List of Formulae (MF19) to show that
$$\sum _ { r = 1 } ^ { n } ( 7 r + 1 ) ( 7 r + 8 ) = a n ^ { 3 } + b n ^ { 2 } + c n$$
where $a , b$ and $c$ are constants to be determined.
\item Use the method of differences to find $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 7 r + 1 ) ( 7 r + 8 ) }$ in terms of $n$.
\item Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( 7 r + 1 ) ( 7 r + 8 ) }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q2 [8]}}