| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Standard summation formulae application |
| Difficulty | Moderate -0.3 This is a straightforward application of standard summation formulas with algebraic manipulation. Part (i) is direct substitution into given formulas. Part (ii) requires splitting the sum, applying formulas, and factorizing to match the given form—routine for FP1 students who know the standard results. The 8 marks reflect length rather than conceptual difficulty. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\sum_{r=1}^{24} (r^3 - 4r) = \frac{1}{4} \cdot 24^2(24+1)^2 - 4 \cdot \frac{1}{2} \cdot 24(24+1)\) | M1 | An attempt to use at least one of the standard formulae correctly and substitute 24. |
| \(\{= 90000 - 1200\} = 88800\) | A1 cao | 88800 |
| (ii) \(\sum_{r=0}^{n} (r^2 - 2r + 2n + 1)\) | ||
| \(= \frac{1}{6}n(n+1)(2n+1) - 2 \cdot \frac{1}{2}n(n+1) + 2n(n+1) + (n+1)\) | M1 | An attempt to use at least one of the standard formulae correctly. Correct underlined expression, \(2n \to 2n(n+1)\), \(1 \to (n+1)\), B1 for \(1 \to (n+1)\), B1 for \(2n \to 2n(n+1)\) |
| \(= \frac{1}{6}n(n+1)(2n+1) - 2 \cdot \frac{1}{2}n(n+1) + 2n(n+1) + (n+1)\) | A1 B1 B1 | An attempt to factorise out \(\frac{1}{6}(n+1)\) or \(\frac{1}{n}\). M1 |
| M1 | ||
| \(= \frac{1}{6}(n+1) \{2n^2 + n - 6n + 12n + 6\}\) | ||
| \(= \frac{1}{6}(n+1)\{2n^2 + n - 6n + 12n + 6\}\) | ||
| \(= \frac{1}{6}(n+1)\{2n^2 + 7n + 6\}\) | ||
| \(= \frac{1}{6}(n+1)(n+2)(2n+3)\) | A1 | Correct answer. (Note: \(a = 2, b = 2, c = 3.\)) |
**(i)** $\sum_{r=1}^{24} (r^3 - 4r) = \frac{1}{4} \cdot 24^2(24+1)^2 - 4 \cdot \frac{1}{2} \cdot 24(24+1)$ | M1 | An attempt to use at least one of the standard formulae correctly and substitute 24.
$\{= 90000 - 1200\} = 88800$ | A1 cao | 88800 | [2]
**(ii)** $\sum_{r=0}^{n} (r^2 - 2r + 2n + 1)$ | |
$= \frac{1}{6}n(n+1)(2n+1) - 2 \cdot \frac{1}{2}n(n+1) + 2n(n+1) + (n+1)$ | M1 | An attempt to use at least one of the standard formulae correctly. Correct underlined expression, $2n \to 2n(n+1)$, $1 \to (n+1)$, B1 for $1 \to (n+1)$, B1 for $2n \to 2n(n+1)$
$= \frac{1}{6}n(n+1)(2n+1) - 2 \cdot \frac{1}{2}n(n+1) + 2n(n+1) + (n+1)$ | A1 B1 B1 | An attempt to factorise out $\frac{1}{6}(n+1)$ or $\frac{1}{n}$. M1
| M1 |
$= \frac{1}{6}(n+1) \{2n^2 + n - 6n + 12n + 6\}$ | |
$= \frac{1}{6}(n+1)\{2n^2 + n - 6n + 12n + 6\}$ | |
$= \frac{1}{6}(n+1)\{2n^2 + 7n + 6\}$ | |
$= \frac{1}{6}(n+1)(n+2)(2n+3)$ | A1 | Correct answer. (Note: $a = 2, b = 2, c = 3.$) | [6]
**Total: [8]**\begin{enumerate}[label=(\roman*)]
\item Use the standard results for $\sum_{r=1}^{n} r^3$ and $\sum_{r=1}^{n} r$ to evaluate
$$\sum_{r=1}^{24} (r^3 - 4r)$$ [2]
\item Use the standard results for $\sum_{r=1}^{n} r^2$ and $\sum_{r=1}^{n} r$ to show that
$$\sum_{r=0}^{n} (r^2 - 2r + 2n + 1) = \frac{1}{6}(n + 1)(n + a)(bn + c)$$
for all integers $n \geqslant 0$, where $a$, $b$ and $c$ are constant integers to be found. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q10 [8]}}