| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Alternating series summation |
| Difficulty | Standard +0.8 This is a Further Maths question requiring manipulation of alternating series with the sum of squares formula. Part (i) demands algebraic insight to group terms and apply the standard formula cleverly. Part (ii) involves limits of rational expressions requiring asymptotic analysis. While systematic, it's above average difficulty due to the alternating series structure and the need for strategic term grouping rather than direct application. |
| Spec | 1.04g Sigma notation: for sums of series4.06b Method of differences: telescoping series8.01c Sequence behaviour: periodic, convergent, divergent, oscillating, monotonic |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{2n} = 1^2 - 2^2 + 3^2 - 4^2 \ldots\) | M1 | Uses correct difference |
| \(S_{2n} = \sum_{r=1}^{2n} r^2 - 2\sum_{r=1}^{n}(2r)^2 = \sum_{r=1}^{2n} r^2 - 8\sum_{r=1}^{n} r^2\) | A1 | Alt method: Use \(\sum_1^n(2r-1)^2 - \sum_1^n(2r)^2 = A1\) |
| \(S_{2n} = \frac{1}{6}(2n)(2n+1)(4n+1) - \frac{8}{6}n(n+1)(2n+1)\) | M1 | \(\sum_1^n 4r^2 - 4\sum_1^n(r) + n - 4\sum_1^n(r)^2\) M1 |
| Factorising, \(S_{2n} = \frac{1}{3}n(2n+1)(4n+1-4n-4) = -n(2n+1)\) | A1 | \(= -n(2n+1)\) A1 AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lim_{n\to\infty} \frac{S_{2n}}{n^2} = -2\) | B1 | |
| \(S_{2n+1} = S_{2n} + (-1)^{2n}(2n+1)^2\) | M1 | |
| \(S_{2n+1} = -n(2n+1) + (2n+1)^2 = (2n+1)(n+1)\) | M1 | Uses result from (i) or using \(\lim_{n\to\infty}\frac{S_{2n}}{n^2}\) and correct sign |
| \(\lim_{n\to\infty} \frac{S_{2n+1}}{n^2} = 2\) | A1 | Alt: Find limit from previous line directly |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{2n} = 1^2 - 2^2 + 3^2 - 4^2 \ldots$ | M1 | Uses correct difference |
| $S_{2n} = \sum_{r=1}^{2n} r^2 - 2\sum_{r=1}^{n}(2r)^2 = \sum_{r=1}^{2n} r^2 - 8\sum_{r=1}^{n} r^2$ | A1 | Alt method: Use $\sum_1^n(2r-1)^2 - \sum_1^n(2r)^2 = A1$ |
| $S_{2n} = \frac{1}{6}(2n)(2n+1)(4n+1) - \frac{8}{6}n(n+1)(2n+1)$ | M1 | $\sum_1^n 4r^2 - 4\sum_1^n(r) + n - 4\sum_1^n(r)^2$ M1 |
| Factorising, $S_{2n} = \frac{1}{3}n(2n+1)(4n+1-4n-4) = -n(2n+1)$ | A1 | $= -n(2n+1)$ A1 AG |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lim_{n\to\infty} \frac{S_{2n}}{n^2} = -2$ | B1 | |
| $S_{2n+1} = S_{2n} + (-1)^{2n}(2n+1)^2$ | M1 | |
| $S_{2n+1} = -n(2n+1) + (2n+1)^2 = (2n+1)(n+1)$ | M1 | Uses result from (i) or using $\lim_{n\to\infty}\frac{S_{2n}}{n^2}$ and correct sign |
| $\lim_{n\to\infty} \frac{S_{2n+1}}{n^2} = 2$ | A1 | Alt: Find limit from previous line directly |5 Let $S _ { n } = \sum _ { r = 1 } ^ { n } ( - 1 ) ^ { r - 1 } r ^ { 2 }$.\\
(i) Use the standard result for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ given in the List of Formulae (MF10) to show that
$$S _ { 2 n } = - n ( 2 n + 1 )$$
(ii) State the value of $\lim _ { n \rightarrow \infty } \frac { S _ { 2 n } } { n ^ { 2 } }$ and find $\lim _ { n \rightarrow \infty } \frac { S _ { 2 n + 1 } } { n ^ { 2 } }$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q5 [8]}}