| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Sums Between Limits |
| Difficulty | Standard +0.8 This is a Further Maths FP1 question requiring manipulation of sum formulas with non-standard limits (r=n to 2n rather than 1 to n). Part (i) requires algebraic manipulation of the standard sum of cubes formula, involving factorization of quartic expressions. Part (ii) requires recognizing how to split and apply the result from (i). While systematic, it demands careful algebra and is above average difficulty due to the non-standard limits and multi-step algebraic manipulation typical of Further Maths. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Difference of sum to \(2n\) and \(n-1\) | M1 | \(\frac{1}{4}(2n)^2(2n+1)^2 - \frac{1}{4}(n-1)^2 n^2\) |
| Correct unsimplified answer | A1 | |
| Sensible attempt to factorise, at least factor \(n^2\) | M1 | |
| Obtain given answer no errors seen | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Difference of (i) and another standard result | M1 | |
| Difference of \(\sum_1^{2n} r - \sum_1^k r\) for \(k = n-1\) or \(n\) | M1 | \(\frac{1}{2}(2n)(2n+1) - \frac{1}{2}n(n-1)\) |
| Obtain complete unsimplified expression | A1 | (i) \(- 2\times\) above |
| Sensible attempt to factorise, at least factor \(n(n+1)\) | M1 | |
| \(\frac{3}{4}n(n+1)^2(5n-4)\) — obtain correct answer | A1 | \((n+1)(n+1)\) is OK for \((n+1)^2\) |
| [5] |
# Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Difference of sum to $2n$ and $n-1$ | M1 | $\frac{1}{4}(2n)^2(2n+1)^2 - \frac{1}{4}(n-1)^2 n^2$ |
| Correct unsimplified answer | A1 | |
| Sensible attempt to factorise, at least factor $n^2$ | M1 | |
| Obtain **given** answer no errors seen | A1 | |
| **[4]** | | |
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# Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Difference of (i) and another standard result | M1 | |
| Difference of $\sum_1^{2n} r - \sum_1^k r$ for $k = n-1$ or $n$ | M1 | $\frac{1}{2}(2n)(2n+1) - \frac{1}{2}n(n-1)$ |
| Obtain complete unsimplified expression | A1 | (i) $- 2\times$ above |
| Sensible attempt to factorise, at least factor $n(n+1)$ | M1 | |
| $\frac{3}{4}n(n+1)^2(5n-4)$ — obtain correct answer | A1 | $(n+1)(n+1)$ is OK for $(n+1)^2$ |
| **[5]** | | |
---8 (i) Show that $\sum _ { r = n } ^ { 2 n } r ^ { 3 } = \frac { 3 } { 4 } n ^ { 2 } ( n + 1 ) ( 5 n + 1 )$.\\
(ii) Hence find $\sum _ { r = n } ^ { 2 n } r \left( r ^ { 2 } - 2 \right)$, giving your answer in a fully factorised form.
\hfill \mbox{\textit{OCR FP1 2014 Q8 [9]}}