| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Marks | 21 |
| Paper | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Standard +0.3 This is a standard FP2 question combining routine partial fractions with method of differences (a well-practiced technique) and a locus problem involving complex numbers. Part (a) is straightforward partial fractions, part (b) is a standard telescoping sum proof, and parts (c)-(d) involve standard circle locus and argument calculations. All techniques are textbook exercises with no novel insight required, making this slightly easier than average for A-level. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines4.05c Partial fractions: extended to quadratic denominators4.06b Method of differences: telescoping series |
\begin{enumerate}[label=(\alph*)]
\item Express $\frac{1}{r(r + 2)}$ in partial fractions. [2]
\item Hence prove, by the method of differences, that
$$\sum_{r=1}^{2n} \frac{1}{r(r + 2)} = \frac{n(4n + 5)}{4(n + 1)(n + 2)},$$ [6]
\end{enumerate}
The point $Q$ represents a complex number $z$ on an Argand diagram such that
$$|z - 6| = 2|z - 3|.$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that, as $z$ varies, the locus of $P$ is a circle, stating the radius and the coordinates of the centre of this circle. [6]
\end{enumerate}
where $a$ and $b$ are constants to be found.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Hence show that
$$\sum_{r=1}^{2n} \frac{1}{r(r + 2)} = \frac{n(4n + 5)}{4(n + 1)(n + 2)},$$ [3]
\item Find the complex number for which both $|z - 6| = 2|z - 3|$ and $\arg(z - 6) = -\frac{3\pi}{4}$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 Q6 [21]}}