| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Geometric series with summation |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear signposting through three parts. Part (a) requires algebraic manipulation using standard summation formulae (routine for FM students). Part (b) combines partial fractions (standard technique) with method of differences applied to a geometric series (moderately challenging but well-practiced). Part (c) is a straightforward limit as Nāā. While requiring multiple techniques, each step is guided and represents standard Further Maths content without requiring novel insight. |
| 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 |
|---|---|---|
| \(\sum_{r=1}^{N} r(r+1)(3r+4) = 3\sum_{r=1}^{N} r^3 + 7\sum_{r=1}^{N} r^2 + 4\sum_{r=1}^{N} r\) | B1 | Expands |
| \(\frac{3}{4}N^2(N+1)^2 + \frac{7}{6}N(N+1)(2N+1) + 2N(N+1)\) | M1 | Substitutes formulae from MF19 |
| \(\frac{1}{12}N\left(9N(N^2+2N+1)+14(2N^2+3N+1)+24N+24\right)\) | A1 | |
| \(\frac{1}{12}N(9N^3+46N^2+75N+38) = \frac{1}{12}N(N+1)(N+2)(9N+19)\) | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3r+4}{r(r+1)} = \frac{4}{r} - \frac{1}{r+1}\) | M1 A1 | Finds partial fractions |
| \(\sum_{r=1}^{N}\frac{3r+4}{r(r+1)}\left(\frac{1}{4}\right)^{r+1} = \left(\frac{1}{4^2}\right)\left(\frac{4}{1}-\frac{1}{2}\right)+\left(\frac{1}{4^3}\right)\left(\frac{4}{2}-\frac{1}{3}\right)+\ldots+\left(\frac{1}{4^{N+1}}\right)\left(\frac{4}{N}-\frac{1}{N+1}\right)\) | M1 | Writes at least three terms, including last |
| \(\frac{1}{4} - \frac{1}{4^{N+1}(N+1)}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{4}\) | B1 | CAO |
## Question 3(a):
| $\sum_{r=1}^{N} r(r+1)(3r+4) = 3\sum_{r=1}^{N} r^3 + 7\sum_{r=1}^{N} r^2 + 4\sum_{r=1}^{N} r$ | B1 | Expands |
| $\frac{3}{4}N^2(N+1)^2 + \frac{7}{6}N(N+1)(2N+1) + 2N(N+1)$ | M1 | Substitutes formulae from MF19 |
| $\frac{1}{12}N\left(9N(N^2+2N+1)+14(2N^2+3N+1)+24N+24\right)$ | A1 | |
| $\frac{1}{12}N(9N^3+46N^2+75N+38) = \frac{1}{12}N(N+1)(N+2)(9N+19)$ | | AG |
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## Question 3(b):
| $\frac{3r+4}{r(r+1)} = \frac{4}{r} - \frac{1}{r+1}$ | M1 A1 | Finds partial fractions |
| $\sum_{r=1}^{N}\frac{3r+4}{r(r+1)}\left(\frac{1}{4}\right)^{r+1} = \left(\frac{1}{4^2}\right)\left(\frac{4}{1}-\frac{1}{2}\right)+\left(\frac{1}{4^3}\right)\left(\frac{4}{2}-\frac{1}{3}\right)+\ldots+\left(\frac{1}{4^{N+1}}\right)\left(\frac{4}{N}-\frac{1}{N+1}\right)$ | M1 | Writes at least three terms, including last |
| $\frac{1}{4} - \frac{1}{4^{N+1}(N+1)}$ | A1 | |
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## Question 3(c):
| $\frac{1}{4}$ | B1 | CAO |
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\begin{enumerate}[label=(\alph*)]
\item Use standard results from the list of formulae (MF19) to show that
$$\sum _ { r = 1 } ^ { N } r ( r + 1 ) ( 3 r + 4 ) = \frac { 1 } { 12 } N ( N + 1 ) ( N + 2 ) ( 9 N + 19 )$$
\item Express $\frac { 3 r + 4 } { r ( r + 1 ) }$ in partial fractions and hence use the method of differences to find
$$\sum _ { r = 1 } ^ { N } \frac { 3 r + 4 } { r ( r + 1 ) } \left( \frac { 1 } { 4 } \right) ^ { r + 1 }$$
in terms of $N$.
\item Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 3 r + 4 } { r ( r + 1 ) } \left( \frac { 1 } { 4 } \right) ^ { r + 1 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q3 [8]}}