| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Sum of Powers Using Standard Formulae |
| Difficulty | Standard +0.3 This is a straightforward application of standard summation formulae with minimal problem-solving required. Part (a) involves routine algebraic manipulation of given formulae, part (b) is immediate from factorization, and part (c) requires subtracting known sums—all mechanical processes typical of FP1. While it's Further Maths content, the question demands only procedural competence rather than insight, making it slightly easier than an average A-level question overall. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^n (r^2 - r - 8)\) | ||
| \(= \frac{1}{6}n(n+1)(2n+1) - \frac{1}{2}n(n+1) - 8n\) | M1 | At least one of the first two terms is correct. Correct expression |
| \(= \frac{1}{6}n((2n+1)(n+1) - 3(n+1) - 48)\) | M1 | An attempt to factorise out at least \(n\). |
| \(= \frac{1}{6}n(2n^2 + 3n + 1 - 3n - 3 - 48)\) | ||
| \(= \frac{1}{6}n(2n^2 - 50)\) | ||
| \(= \frac{2}{6}n(n^2 - 25)\) | ||
| \(= \frac{1}{3}n(n-5)(n+5)\) | A1 | Achieves the correct answer. |
| Answer | Marks | Guidance |
|---|---|---|
| \(n = 5\) | B1 cao | Give B0 for 2 or more possible values of \(n\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{k}{4}(17)^2(8)^2 - \frac{k}{4}(3)^2(2)^2\right) + \left(\frac{1}{3}(17)(22)(12) - \frac{1}{3}(2)(-3)(7)\right)\) | M1 | Applying at least one of \(n = 17\) or \(n = 2\) to both \(\frac{1}{4}n^2(n+1)^2\) and their \(\frac{1}{3}n(n-5)(n+5)\) |
| M1 | Applying \(n = 17\) and \(n = 2\) only to both \(\frac{1}{4}n^2(n+1)^2\) and their \(\frac{1}{3}n(n-5)(n+5)\). Require differences only for both brackets. | |
| \(\{\Sigma = 6710 \Rightarrow 23409k - 9k + 1496 + 14 = 6710 \Rightarrow k = \frac{2}{9}\) | ddM1 | Sets their sum to 6710 and solves to give \(k = ...\) |
| A1 cso | \(k = \frac{2}{9}\) or \(0.2\) |
**Part (a)**
| $\sum_{r=1}^n (r^2 - r - 8)$ | | |
| $= \frac{1}{6}n(n+1)(2n+1) - \frac{1}{2}n(n+1) - 8n$ | M1 | At least one of the first two terms is correct. Correct expression |
| $= \frac{1}{6}n((2n+1)(n+1) - 3(n+1) - 48)$ | M1 | An attempt to factorise out at least $n$. |
| $= \frac{1}{6}n(2n^2 + 3n + 1 - 3n - 3 - 48)$ | | |
| $= \frac{1}{6}n(2n^2 - 50)$ | | |
| $= \frac{2}{6}n(n^2 - 25)$ | | |
| $= \frac{1}{3}n(n-5)(n+5)$ | A1 | Achieves the correct answer. |
**Part (b)**
| $n = 5$ | B1 cao | Give B0 for 2 or more possible values of $n$. |
**Part (c)**
| $\left(\frac{k}{4}(17)^2(8)^2 - \frac{k}{4}(3)^2(2)^2\right) + \left(\frac{1}{3}(17)(22)(12) - \frac{1}{3}(2)(-3)(7)\right)$ | M1 | Applying at least one of $n = 17$ or $n = 2$ to both $\frac{1}{4}n^2(n+1)^2$ and their $\frac{1}{3}n(n-5)(n+5)$ |
| | M1 | Applying $n = 17$ and $n = 2$ only to both $\frac{1}{4}n^2(n+1)^2$ and their $\frac{1}{3}n(n-5)(n+5)$. Require differences only for both brackets. |
| $\{\Sigma = 6710 \Rightarrow 23409k - 9k + 1496 + 14 = 6710 \Rightarrow k = \frac{2}{9}$ | ddM1 | Sets their sum to 6710 and solves to give $k = ...$ |
| | A1 cso | $k = \frac{2}{9}$ or $0.2$ |
**Total: 9 marks**
---\begin{enumerate}
\item (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that, for all positive integers $n$,
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \left( r ^ { 2 } - r - 8 \right) = \frac { 1 } { 3 } n ( n - a ) ( n + a )$$
where $a$ is a positive integer to be determined.\\
(b) Hence, or otherwise, state the positive value of $n$ that satisfies
$$\sum _ { r = 1 } ^ { n } \left( r ^ { 2 } - r - 8 \right) = 0$$
Given that
$$\sum _ { r = 3 } ^ { 17 } \left( k r ^ { 3 } + r ^ { 2 } - r - 8 \right) = 6710 \quad \text { where } k \text { is a constant }$$
(c) find the exact value of $k$.\\
\hfill \mbox{\textit{Edexcel FP1 2018 Q4 [9]}}