| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Sums Between Limits |
| Difficulty | Standard +0.3 This is a straightforward application of standard summation formulas. Part (a) requires algebraic manipulation of known results for Σr and Σr³, which is routine for FP1 students. Part (b) applies the difference formula Σ(r=10 to 50) = Σ(r=1 to 50) - Σ(r=1 to 9), requiring only substitution and arithmetic. While it involves multiple steps, no novel insight is needed—it's a textbook exercise testing formula recall and careful algebra, making it slightly easier than average. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r(r^2-3) = r^3 - 3r\) | B1 | \(r^3 - 3r\) |
| \(\sum_{r=1}^{n} r(r^2-3) = \sum_{r=1}^{n} r^3 - 3\sum_{r=1}^{n} r = \frac{1}{4}n^2(n+1)^2 - \frac{3}{2}n(n+1)\) | M1A1 | M1: attempt to use at least one standard formula correctly; A1: correct expression |
| \(= \frac{1}{4}n(n+1)\bigl(n(n+1)-6\bigr)\) | M1 | Attempt to factor \(\frac{1}{4}n(n+1)\) before given answer |
| \(= \frac{1}{4}n(n+1)(n^2+n-6) = \frac{1}{4}n(n+1)(n+3)(n-2)\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=10}^{50} r(r^2-3) = f(50) - f(9\ \text{or}\ 10)\) | M1 | Require some use of the result in part (a) for method |
| \(= \frac{1}{4}(50)(51)(53)(48) - \frac{1}{4}(9)(10)(12)(7)\) | A1 | Correct expression |
| \(= 1621800 - 1890 = 1619910\) | A1 | cao |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r(r^2-3) = r^3 - 3r$ | B1 | $r^3 - 3r$ |
| $\sum_{r=1}^{n} r(r^2-3) = \sum_{r=1}^{n} r^3 - 3\sum_{r=1}^{n} r = \frac{1}{4}n^2(n+1)^2 - \frac{3}{2}n(n+1)$ | M1A1 | M1: attempt to use at least one standard formula correctly; A1: correct expression |
| $= \frac{1}{4}n(n+1)\bigl(n(n+1)-6\bigr)$ | M1 | Attempt to factor $\frac{1}{4}n(n+1)$ before given answer |
| $= \frac{1}{4}n(n+1)(n^2+n-6) = \frac{1}{4}n(n+1)(n+3)(n-2)$ | A1 | cso |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=10}^{50} r(r^2-3) = f(50) - f(9\ \text{or}\ 10)$ | M1 | Require some use of the result in part (a) for method |
| $= \frac{1}{4}(50)(51)(53)(48) - \frac{1}{4}(9)(10)(12)(7)$ | A1 | Correct expression |
| $= 1621800 - 1890 = 1619910$ | A1 | cao |\begin{enumerate}
\item (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ to show that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } r \left( r ^ { 2 } - 3 \right) = \frac { 1 } { 4 } n ( n + 1 ) ( n + 3 ) ( n - 2 )$$
(b) Calculate the value of $\sum _ { r = 10 } ^ { 50 } r \left( r ^ { 2 } - 3 \right)$\\
\hfill \mbox{\textit{Edexcel FP1 2014 Q5 [8]}}