| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Finding constants from given sum formula |
| Difficulty | Standard +0.3 This question requires applying standard summation formulas (Σr², Σr³) and algebraic manipulation to verify a given result, then solving for a constant by equating coefficients. While it involves multiple steps and careful algebra, it's a straightforward application of well-practiced techniques from the FP1 syllabus with no novel problem-solving required—slightly easier than average for Further Maths. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks |
|---|---|
| \(\sum r^2(2r-p) = 2\sum r^3 - p\sum r^2\) | M1 |
| \(= 2\cdot\frac{n^2(n+1)^2}{4} - p\cdot\frac{n(n+1)(2n+1)}{6}\) | M1 A1 |
| \(= \frac{1}{6}n(n+1)(3n^2+(3-2p)n-p)\) | A1 |
| Answer | Marks |
|---|---|
| Coefficient of \(n^3\): \(3\), coefficient of \(n^4\): \(\frac{1}{2}\) (from expansion) | M1 |
| Setting equal: \(3-2p = 3 \Rightarrow p = 0\)... comparing correctly gives \(p = \frac{3}{2}\) | A1 |
## Question 4:
**(i)**
$\sum r^2(2r-p) = 2\sum r^3 - p\sum r^2$ | M1 |
$= 2\cdot\frac{n^2(n+1)^2}{4} - p\cdot\frac{n(n+1)(2n+1)}{6}$ | M1 A1 |
$= \frac{1}{6}n(n+1)(3n^2+(3-2p)n-p)$ | A1 |
**(ii)**
Coefficient of $n^3$: $3$, coefficient of $n^4$: $\frac{1}{2}$ (from expansion) | M1 |
Setting equal: $3-2p = 3 \Rightarrow p = 0$... comparing correctly gives $p = \frac{3}{2}$ | A1 |
---4 (i) Use standard series to show that
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 2 r - p ) = \frac { 1 } { 6 } n ( n + 1 ) \left( 3 n ^ { 2 } + ( 3 - 2 p ) n - p \right) ,$$
where $p$ is a constant.\\
(ii) Given that the coefficients of $n ^ { 3 }$ and $n ^ { 4 }$ in the expression for $\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 2 r - p )$ are equal, find the value of $p$.
\hfill \mbox{\textit{OCR MEI FP1 2016 Q4 [6]}}