| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Standard summation formulae application |
| Difficulty | Standard +0.3 This is a straightforward proof by induction (or algebraic manipulation) followed by a simple deduction about divisibility. Part (a) requires standard summation techniques with a given result to prove, and part (b) is a direct consequence requiring minimal additional reasoning. While it involves multiple steps, the techniques are routine for Further Maths students and the result is provided rather than requiring discovery. |
| Spec | 4.01a Mathematical induction: construct proofs8.02l Fermat's little theorem: both forms |
\begin{enumerate}[label=(\alph*)]
\item Prove that $\sum_{r=1}^{n} (r + 1)(r - 1) = \frac{1}{6} n (n - 1)(2n + 5)$. [5]
\item Deduce that $n(n - 1)(2n + 5)$ is divisible by 6 for all $n > 1$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 Q2 [7]}}