| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Partial fractions then method of differences |
| Difficulty | Standard +0.8 This is a standard Further Maths question combining partial fractions with method of differences, but requires careful algebraic manipulation across three connected parts. Part (a) uses formula booklet results, part (b) is routine partial fractions followed by telescoping series, and part (c) requires understanding limits. The multi-step nature and need to connect all parts elevates it slightly above average Further Maths difficulty, but it follows well-established techniques 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(\displaystyle\sum_{r=1}^{n}(r^3 + 3r^2 + 2r) = \tfrac{1}{4}n^2(n+1)^2 + \tfrac{1}{2}n(n+1)(2n+1) + n(n+1)\) | M1 A1 | Expands and substitutes formulae. At least two formulae used for M1. |
| \(= \tfrac{1}{4}n(n+1)\!\left(n^2 + n + 4n + 2 + 4\right) = \tfrac{1}{4}n(n+1)(n+2)(n+3)\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{r(r+1)(r+2)} = \frac{1}{2r} - \frac{1}{r+1} + \frac{1}{2(r+2)}\) | M1 A1 | Finds partial fractions |
| \(\sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)} = \frac{1}{2}f(1) - f(2) + \frac{1}{2}f(3)\) \(+\frac{1}{2}f(2) - f(3) + \frac{1}{2}f(4)\) \(+\frac{1}{2}f(3) - f(4) + \frac{1}{2}f(5)\) \(\vdots\) \(+\frac{1}{2}f(n-1) - f(n) + \frac{1}{2}f(n+1)\) \(+\frac{1}{2}f(n) - f(n+1) + \frac{1}{2}f(n+2)\) | M1 A1 | Shows enough terms for cancellation to be clear. Where \(f(x) = \frac{1}{r}\) |
| \(= \frac{1}{4} - \frac{1}{2(n+1)} + \frac{1}{2(n+2)}\) | A1 | ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{4}\) | B1 FT | FT from *their* answer to part (b) |
**Question 2(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\displaystyle\sum_{r=1}^{n}(r^3 + 3r^2 + 2r) = \tfrac{1}{4}n^2(n+1)^2 + \tfrac{1}{2}n(n+1)(2n+1) + n(n+1)$ | M1 A1 | Expands and substitutes formulae. At least two formulae used for M1. |
| $= \tfrac{1}{4}n(n+1)\!\left(n^2 + n + 4n + 2 + 4\right) = \tfrac{1}{4}n(n+1)(n+2)(n+3)$ | A1 | |
| **Total** | **3** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{r(r+1)(r+2)} = \frac{1}{2r} - \frac{1}{r+1} + \frac{1}{2(r+2)}$ | M1 A1 | Finds partial fractions |
| $\sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)} = \frac{1}{2}f(1) - f(2) + \frac{1}{2}f(3)$ $+\frac{1}{2}f(2) - f(3) + \frac{1}{2}f(4)$ $+\frac{1}{2}f(3) - f(4) + \frac{1}{2}f(5)$ $\vdots$ $+\frac{1}{2}f(n-1) - f(n) + \frac{1}{2}f(n+1)$ $+\frac{1}{2}f(n) - f(n+1) + \frac{1}{2}f(n+2)$ | M1 A1 | Shows enough terms for cancellation to be clear. Where $f(x) = \frac{1}{r}$ |
| $= \frac{1}{4} - \frac{1}{2(n+1)} + \frac{1}{2(n+2)}$ | A1 | ISW |
## Question 2(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{4}$ | B1 FT | FT from *their* answer to part (b) |2
\begin{enumerate}[label=(\alph*)]
\item Use standard results from the list of formulae (MF19) to find $\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 2 )$ in terms of $n$,\\
fully factorising your answer. fully factorising your answer.
\item Express $\frac { 1 } { r ( r + 1 ) ( r + 2 ) }$ in partial fractions and hence use the method of differences to find
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$$
\item Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q2 [9]}}