| Exam Board | Edexcel |
|---|---|
| Module | AEA (Advanced Extension Award) |
| Year | 2011 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Geometric series with summation |
| Difficulty | Challenging +1.8 This AEA question requires recognizing a pattern in alternating recurrence relations, deriving a geometric series sum formula through careful algebraic manipulation, and applying the floor function creatively to match the general form. While systematic, it demands strong pattern recognition, proof technique, and the ability to connect abstract sequences to concrete series—significantly above standard A-level but accessible to well-prepared students. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series1.04i Geometric sequences: nth term and finite series sum |
A sequence $\{u_n\}$ is given by
$$u_1 = k$$
$$u_{2n} = u_{2n-1} \times p \qquad n \geq 1$$
$$u_{2n+1} = u_{2n} \times q \qquad n \geq 1$$
where $k$, $p$ and $q$ are positive constants with $pq \neq 1$
\begin{enumerate}[label=(\alph*)]
\item Write down the first 6 terms of this sequence.
[3]
\item Show that $\sum_{r=1}^{2n} u_r = \frac{k(1+p)(1-(pq)^n)}{1-pq}$
[6]
\end{enumerate}
In part (c) $[x]$ means the integer part of $x$, so for example $[2.73] = 2$, $[4] = 4$ and $[0] = 0$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find $\sum_{r=1}^{\infty} 6 \times \left(\frac{4}{3}\right)^{\left[\frac{r}{2}\right]} \times \left(\frac{3}{5}\right)^{\left[\frac{r-1}{2}\right]}$
[4]
\end{enumerate}
[Total 13 marks]
\hfill \mbox{\textit{Edexcel AEA 2011 Q3 [17]}}