| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Proving standard summation formulae |
| Difficulty | Challenging +1.2 Part (a) is a standard method of differences proof for sum of squares that appears frequently in Further Maths syllabi. Parts (b) and (c) require pattern recognition in an alternating series and algebraic manipulation, but follow directly from part (a). The multi-part structure and need to handle the alternating coefficient pattern elevates this slightly above average difficulty, but it remains a fairly routine Further Maths question with well-signposted steps. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(8r^3 + 12r^2 + 6r + 1 - (8r^3 - 12r^2 + 6r - 1) = 24r^2 + 2\) | M1 | Expands |
| \(24\displaystyle\sum_{r=1}^{n} r^2 + 2n = (2n+1)^3 - 1\) | M1 A1 A1 | Sums both sides and uses method of differences with sufficient complete terms |
| \(24\displaystyle\sum_{r=1}^{n} r^2 = 8n^3 + 12n^2 + 4n = 4n(n+1)(2n+1)\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{2n} = \displaystyle\sum_{r=1}^{2n} r^2 + 2\sum_{r=1}^{n}(2r)^2 = \sum_{r=1}^{2n} r^2 + 8\sum_{r=1}^{n} r^2\) | M1 | Relates with sum of squares |
| \(S_{2n} = \frac{1}{6}(2n)(2n+1)(4n+1) + \frac{8}{6}n(n+1)(2n+1)\) | M1 | Applies \(\displaystyle\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\) |
| \(S_{2n} = \frac{1}{3}n(2n+1)(4n+1+4n+4) = \frac{1}{3}n(2n+1)(8n+5)\) | A1 | \(\displaystyle\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{2n} = \displaystyle\sum_{r=1}^{n}(2r-1)^2 + 3\sum_{r=1}^{n}(2r)^2 = \sum_{r=1}^{n}(4r^2 - 4r + 1 + 12r^2)\) | M1 | Relates with sum of squares |
| \(\dfrac{16n(n+1)(2n+1)}{6} - 4\dfrac{n(n+1)}{2} + n\) | M1 | Applies \(\displaystyle\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\) |
| \(\dfrac{n}{3}(16n^2 + 18n + 5) = \dfrac{n}{3}(2n+1)(8n+5)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{16}{3}\) | B1 FT |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $8r^3 + 12r^2 + 6r + 1 - (8r^3 - 12r^2 + 6r - 1) = 24r^2 + 2$ | M1 | Expands |
| $24\displaystyle\sum_{r=1}^{n} r^2 + 2n = (2n+1)^3 - 1$ | M1 A1 A1 | Sums both sides and uses method of differences with sufficient complete terms |
| $24\displaystyle\sum_{r=1}^{n} r^2 = 8n^3 + 12n^2 + 4n = 4n(n+1)(2n+1)$ | A1 | AG |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{2n} = \displaystyle\sum_{r=1}^{2n} r^2 + 2\sum_{r=1}^{n}(2r)^2 = \sum_{r=1}^{2n} r^2 + 8\sum_{r=1}^{n} r^2$ | M1 | Relates with sum of squares |
| $S_{2n} = \frac{1}{6}(2n)(2n+1)(4n+1) + \frac{8}{6}n(n+1)(2n+1)$ | M1 | Applies $\displaystyle\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)$ |
| $S_{2n} = \frac{1}{3}n(2n+1)(4n+1+4n+4) = \frac{1}{3}n(2n+1)(8n+5)$ | A1 | $\displaystyle\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)$ |
**Alternative for 3(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{2n} = \displaystyle\sum_{r=1}^{n}(2r-1)^2 + 3\sum_{r=1}^{n}(2r)^2 = \sum_{r=1}^{n}(4r^2 - 4r + 1 + 12r^2)$ | M1 | Relates with sum of squares |
| $\dfrac{16n(n+1)(2n+1)}{6} - 4\dfrac{n(n+1)}{2} + n$ | M1 | Applies $\displaystyle\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)$ and $\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)$ |
| $\dfrac{n}{3}(16n^2 + 18n + 5) = \dfrac{n}{3}(2n+1)(8n+5)$ | A1 | |
---
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{16}{3}$ | B1 FT | |
---3
\begin{enumerate}[label=(\alph*)]
\item By considering $( 2 r + 1 ) ^ { 3 } - ( 2 r - 1 ) ^ { 3 }$, use the method of differences to prove that
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$
Let $S _ { n } = 1 ^ { 2 } + 3 \times 2 ^ { 2 } + 3 ^ { 2 } + 3 \times 4 ^ { 2 } + 5 ^ { 2 } + 3 \times 6 ^ { 2 } + \ldots + \left( 2 + ( - 1 ) ^ { n } \right) n ^ { 2 }$.
\item Show that $\mathrm { S } _ { 2 \mathrm { n } } = \frac { 1 } { 3 } \mathrm { n } ( 2 \mathrm { n } + 1 ) ( \mathrm { an } + \mathrm { b } )$, where $a$ and $b$ are integers to be determined.
\item State the value of $\lim _ { n \rightarrow \infty } \frac { S _ { 2 n } } { n ^ { 3 } }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q3 [9]}}