| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2004 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Proving standard summation formulae |
| Difficulty | Standard +0.3 This is a standard method-of-differences proof taught explicitly in FP2. Part (a) is routine algebra expanding and collecting terms. Part (b) follows a well-rehearsed technique: sum the identity from r=1 to n, telescope the left side, and rearrange. While it requires careful algebraic manipulation, it's a textbook example with no novel insight needed, making it slightly easier than average. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
(b) $AB = 2r\sin\theta$ M1 A1
Area $= \int_{-\pi/3}^{\pi/3} \frac{1}{2}r^2 d\theta$ M1
$= \frac{1}{2a^2}\int[a^2(1+\cos\theta)^2 - 9a^2(1-\cos\theta)^2] d\theta$ M1
$= \frac{1}{2a^2}\int[1+2\cos\theta+\cos^2\theta - 9(1-2\cos\theta+\cos^2\theta)] d\theta$ A1
$= \frac{1}{2a^2}\int[-8+20\cos\theta-8\cos^2\theta] d\theta$
$= k[-8\theta + 20\sin\theta \ldots$ B1
$\ldots - 2\sin 2\theta - 4\theta]$ B1
Uses limits $\frac{\pi}{3}$ and $-\frac{\pi}{3}$ correctly or uses twice smaller area and uses limits $\frac{\pi}{3}$ and $0$ correctly. (Need not see $0$ substituted)
$= a^2[-4\pi + 10\sqrt{3} - 3]$ or $= a^2[-4\pi + 9\sqrt{3}]$ or $3.022a^2$ A1
(d) $3a = 4.5 \rightarrow a = \frac{3}{2}$ B1
$\therefore \text{Area} = 3[9\sqrt{3} - 4\pi] = 9.07 \text{ cm}^2$ M1 A1
---\begin{enumerate}[label=(\alph*)]
\item Show that $( r + 1 ) ^ { 3 } - ( r - 1 ) ^ { 3 } \equiv A r ^ { 2 } + B$, where $A$ and $B$ are constants to be found.
\item Prove by the method of differences that $\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 ) , n > 1$.\\
(6)(Total 8 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2004 Q1 [8]}}