| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | 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 straightforward proof by induction followed by algebraic manipulation. Part (a) is a standard induction exercise with routine base case, inductive step, and algebraic simplification. Part (b) requires recognizing that the sum from n+1 to 2n equals S(2n) - S(n), then factoring the result—mechanical algebra with no novel insight required. Slightly easier than average due to its formulaic nature. |
| Spec | 4.01a Mathematical induction: construct proofs4.06b Method of differences: telescoping series |
\begin{enumerate}[label=(\alph*)]
\item Prove by induction, that for $n \in \mathbb{Z}^+$,
$$\sum_{r=1}^{n} r(2r - 1) = \frac{1}{6}n(n + 1)(4n - 1)$$ [6]
\item Hence, show that
$$\sum_{r=n+1}^{2n} r(2r - 1) = \frac{1}{3}n(an^2 + bn + c)$$
where $a$, $b$ and $c$ are integers to be found. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q8 [10]}}