| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | December |
| Marks | 7 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring partial fractions decomposition and telescoping series techniques. While it involves multiple steps (partial fractions, finding the sum pattern, taking the limit), these are standard Further Pure techniques with no novel insight required. The telescoping pattern becomes clear once partial fractions are applied, making this a moderately challenging but routine Further Maths exercise. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{1}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2}\); \(\Rightarrow A(r+1)(r+2) + Br(r+2) + Cr(r+1) = 1\); \(\Rightarrow A = \frac{1}{2}, B = -1, C = \frac{1}{2}\); \(\Rightarrow \frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{2}{r+1} + \frac{1}{r+2}\right)\) | M1, M1, A1 | Attempt to find partial fractions; Attempt to find constants by comparing 3 coefficients or substituting 3 values for \(r\) |
| \(\sum_{r=1}^n \frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left[\left(\frac{1}{1} - \frac{2}{2} + \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right) + ... + \left(\frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2}\right)\right]\) | M1, M1 | Expand sum using their partial fractions; Cancel terms |
| \(= \frac{1}{2}\left[\left(\frac{1}{1} - \frac{2}{2}\right) + \left(-\frac{1}{n+1} + \frac{1}{n+2}\right)\right] = \frac{1}{2}\left[\frac{1}{1} - \frac{2}{2} - \left(-\frac{1}{n+1} + \frac{1}{n+2}\right)\right]\) | A1 | Answer in the form \(\frac{1}{4} - \frac{1}{2}f(n)\) |
| \(= \frac{1}{4} - \frac{1}{2(n+1)} + \frac{1}{2(n+2)}\) | A1 | |
| (b) \(\frac{1}{4}\) | B1 |
**(a)** $\frac{1}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2}$; $\Rightarrow A(r+1)(r+2) + Br(r+2) + Cr(r+1) = 1$; $\Rightarrow A = \frac{1}{2}, B = -1, C = \frac{1}{2}$; $\Rightarrow \frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{2}{r+1} + \frac{1}{r+2}\right)$ | M1, M1, A1 | Attempt to find partial fractions; Attempt to find constants by comparing 3 coefficients or substituting 3 values for $r$
$\sum_{r=1}^n \frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left[\left(\frac{1}{1} - \frac{2}{2} + \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right) + ... + \left(\frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2}\right)\right]$ | M1, M1 | Expand sum using their partial fractions; Cancel terms
$= \frac{1}{2}\left[\left(\frac{1}{1} - \frac{2}{2}\right) + \left(-\frac{1}{n+1} + \frac{1}{n+2}\right)\right] = \frac{1}{2}\left[\frac{1}{1} - \frac{2}{2} - \left(-\frac{1}{n+1} + \frac{1}{n+2}\right)\right]$ | A1 | Answer in the form $\frac{1}{4} - \frac{1}{2}f(n)$
$= \frac{1}{4} - \frac{1}{2(n+1)} + \frac{1}{2(n+2)}$ | A1 |
**(b)** $\frac{1}{4}$ | B1 |
---7
\begin{enumerate}[label=(\alph*)]
\item Determine an expression for $\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$ giving your answer in the form $\frac { 1 } { 4 } - \frac { 1 } { 2 } \mathrm { f } ( n )$.
\item Find the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q7 [7]}}