| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation with fractions |
| Difficulty | Challenging +1.2 This is a Core Pure 1 induction proof requiring students to find constants a, b, c and prove the result. While it involves algebraic fractions and requires careful manipulation in the inductive step, it follows a standard induction template with no particularly novel insights needed. The partial fraction decomposition is straightforward, and the algebraic simplification, though requiring attention to detail, is routine for Further Maths students. Slightly above average difficulty due to the multi-parameter nature and algebraic complexity, but well within expected CP1 scope. |
| Spec | 4.01a Mathematical induction: construct proofs4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{1}{(r+1)(r+2)(r+3)} \equiv \frac{A}{r+1} + \frac{B}{r+2} + \frac{C}{r+3} \Rightarrow A = \frac{1}{2},\ B = -1,\ C = \frac{1}{2}\) | M1 | 3.1a |
| Writing out terms for \(r=0,\ r=1,\ r=n-1,\ r=n\) e.g. \(r=0\): \(\frac{1}{2}\left[\frac{1}{1} - \frac{2}{2} + \frac{1}{3}\right]\); \(r=n\): \(\frac{1}{2}\left[\frac{1}{n+1} - \frac{2}{n+2} + \frac{1}{n+3}\right]\) | M1 | 2.1 |
| \(\frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{1}{2(n+2)} - \frac{1}{n+2} + \frac{1}{2(n+3)}\) or \(\frac{1}{4} - \frac{1}{2(n+2)} + \frac{1}{2(n+3)}\) | A1 | 1.1b |
| \(= \frac{n^2 + 5n + 6 + 2n + 6 - 4n - 12 + 2n + 4}{4(n+2)(n+3)}\) | M1 | 1.1b |
| \(= \frac{(n+1)(n+4)}{4(n+2)(n+3)}\) | A1 | 2.2a |
# Question 4:
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{1}{(r+1)(r+2)(r+3)} \equiv \frac{A}{r+1} + \frac{B}{r+2} + \frac{C}{r+3} \Rightarrow A = \frac{1}{2},\ B = -1,\ C = \frac{1}{2}$ | M1 | 3.1a |
| Writing out terms for $r=0,\ r=1,\ r=n-1,\ r=n$ e.g. $r=0$: $\frac{1}{2}\left[\frac{1}{1} - \frac{2}{2} + \frac{1}{3}\right]$; $r=n$: $\frac{1}{2}\left[\frac{1}{n+1} - \frac{2}{n+2} + \frac{1}{n+3}\right]$ | M1 | 2.1 |
| $\frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{1}{2(n+2)} - \frac{1}{n+2} + \frac{1}{2(n+3)}$ or $\frac{1}{4} - \frac{1}{2(n+2)} + \frac{1}{2(n+3)}$ | A1 | 1.1b |
| $= \frac{n^2 + 5n + 6 + 2n + 6 - 4n - 12 + 2n + 4}{4(n+2)(n+3)}$ | M1 | 1.1b |
| $= \frac{(n+1)(n+4)}{4(n+2)(n+3)}$ | A1 | 2.2a |
**(5 marks)**
---\begin{enumerate}
\item Prove that, for $n \in \mathbb { Z } , n \geqslant 0$
\end{enumerate}
$$\sum _ { r = 0 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) } = \frac { ( n + a ) ( n + b ) } { c ( n + 2 ) ( n + 3 ) }$$
where $a$, $b$ and $c$ are integers to be found.
\hfill \mbox{\textit{Edexcel CP1 2019 Q4 [5]}}