| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Moderate -0.3 This is a standard proof by induction with a straightforward algebraic step. The partial fractions decomposition of 1/(r(r+1)) = 1/r - 1/(r+1) makes the inductive step telescoping and routine. While it's a Further Maths question, it requires only basic induction technique with no conceptual challenges, making it slightly easier than average overall. |
| Spec | 4.01a Mathematical induction: construct proofs |
Prove by induction that, for $n \in \mathbb{Z}^+$,
$$\sum_{r=1}^n \frac{1}{r(r+1)} = \frac{n}{n+1}$$
[5]
\hfill \mbox{\textit{Edexcel FP1 Q4 [5]}}