| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Challenging +1.2 Part (a) is routine application of standard formulae. Part (b) requires verifying a given identity (algebraic manipulation) then applying method of differences, which is a standard Further Maths technique. Part (c) is a straightforward limit as nāā. The identity is provided rather than requiring students to discover it, making this a methodical rather than insightful question. Slightly above average due to the multi-step nature and Further Maths content, but all techniques are standard. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(n - \frac{1}{2}n(n+1) - \frac{1}{6}n(n+1)(2n+1)\) | M1 A1 | Substitutes correct formulae from MF19; expanding brackets correctly |
| \(\frac{1}{3}n - n^2 - \frac{1}{3}n^3\) | A1 | Simplifies by collecting terms |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{r+1}{(r+1)^2+1} - \frac{r}{r^2+1} = \frac{(r+1)(r^2+1)-r(r^2+2r+2)}{(r^2+2r+2)(r^2+1)} = \frac{1-r-r^2}{(r^2+2r+2)(r^2+1)}\) | M1 A1 | Puts over a common denominator and expands, AG |
| \(\sum_{r=1}^{n} \frac{1-r-r^2}{(r^2+2r+2)(r^2+1)} = \sum_{r=1}^{n}\left(\frac{r+1}{(r+1)^2+1} - \frac{r}{r^2+1}\right) = \frac{2}{5}-\frac{1}{2}+\frac{3}{10}-\frac{2}{5}+\cdots+\frac{n+1}{(n+1)^2+1}-\frac{n}{n^2+1}\) | M1 A1 | Shows at least three complete terms including first and last. Cancellation may be implicit |
| \(= -\frac{1}{2} + \frac{n+1}{(n+1)^2+1}\) | A1 | ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-\frac{1}{2}\) | B1 FT | FT from their answer to part (b) |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n - \frac{1}{2}n(n+1) - \frac{1}{6}n(n+1)(2n+1)$ | **M1 A1** | Substitutes correct formulae from MF19; expanding brackets correctly |
| $\frac{1}{3}n - n^2 - \frac{1}{3}n^3$ | **A1** | Simplifies by collecting terms |
| **Total** | **3** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{r+1}{(r+1)^2+1} - \frac{r}{r^2+1} = \frac{(r+1)(r^2+1)-r(r^2+2r+2)}{(r^2+2r+2)(r^2+1)} = \frac{1-r-r^2}{(r^2+2r+2)(r^2+1)}$ | M1 A1 | Puts over a common denominator and expands, AG |
| $\sum_{r=1}^{n} \frac{1-r-r^2}{(r^2+2r+2)(r^2+1)} = \sum_{r=1}^{n}\left(\frac{r+1}{(r+1)^2+1} - \frac{r}{r^2+1}\right) = \frac{2}{5}-\frac{1}{2}+\frac{3}{10}-\frac{2}{5}+\cdots+\frac{n+1}{(n+1)^2+1}-\frac{n}{n^2+1}$ | M1 A1 | Shows at least three complete terms including first and last. Cancellation may be implicit |
| $= -\frac{1}{2} + \frac{n+1}{(n+1)^2+1}$ | A1 | ISW |
## Question 2(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-\frac{1}{2}$ | 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 $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \left( 1 - \mathrm { r } - \mathrm { r } ^ { 2 } \right)$ in terms of $n$,\\
simplifying your answer. simplifying your answer.
\item Show that
$$\frac { 1 - r - r ^ { 2 } } { \left( r ^ { 2 } + 2 r + 2 \right) \left( r ^ { 2 } + 1 \right) } = \frac { r + 1 } { ( r + 1 ) ^ { 2 } + 1 } - \frac { r } { r ^ { 2 } + 1 }$$
and hence use the method of differences to find $\sum _ { r = 1 } ^ { n } \frac { 1 - r - r ^ { 2 } } { \left( r ^ { 2 } + 2 r + 2 \right) \left( r ^ { 2 } + 1 \right) }$.
\item Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 - r - r ^ { 2 } } { \left( r ^ { 2 } + 2 r + 2 \right) \left( r ^ { 2 } + 1 \right) }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q2 [9]}}