| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Standard +0.8 This is a Further Maths FP1 question requiring expansion of a quartic summation, manipulation using standard summation formulas (∑r, ∑r², ∑r³), algebraic factorization to match a given form, then applying the telescoping technique (sum from 11 to 20 = sum to 20 minus sum to 10). While mechanical, it involves multiple steps with quartic algebra and is more demanding than typical A-level Core questions. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 4\displaystyle\sum_{r=1}^{n}r^3 - 6\sum_{r=1}^{n}r^2 - 2\sum_{r=1}^{n}r\) | M1 | Splitting the sum into three separate sums. PI by m1 line below. |
| \(= 4\times\frac{1}{4}n^2(n+1)^2 - 6\times\frac{1}{6}n(n+1)(2n+1) - 2\times\frac{1}{2}n(n+1)\) | m1 | Substitution of the three summations from FB into \(a\sum r^3 + b\sum r^2 + c\sum r\) |
| \(= n^2(n+1)^2 - n(n+1)(2n+1) - n(n+1)\) | A1 | PI by later expressions |
| \(= n(n+1)[n(n+1)-(2n+1)-1]\) | m1 | Taking out factor \(n(n+1)\) from correct expressions |
| \(= n(n+1)[n^2-n-2]\) | A1 | |
| \(= n(n+1)(n+1)(n-2)\ \left(= n(n-2)(n+1)^2\ (p=-2, q=1)\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\displaystyle\sum_{r=11}^{20} 2r(2r^2-3r-1) = \sum_{r=1}^{20} 2r(2r^2-3r-1) - \sum_{r=1}^{10} 2r(2r^2-3r-1)\) | M1 | \(\sum_{r=1}^{20}\cdots - \sum_{r=1}^{10}\cdots\). PI by next line (ft c's \(p\) & \(q\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 20\times18\times21^2 - 10\times8\times11^2 = 158760 - 9680 = 149080\) | A1 | NMS 0/2. A0 if not showing use of fully factorised form. |
## Question 8:
### Part (a):
$\displaystyle\sum_{r=1}^{n} 2r(2r^2-3r-1) = \sum_{r=1}^{n}4r^3 - \sum_{r=1}^{n}6r^2 - \sum_{r=1}^{n}2r$
$= 4\displaystyle\sum_{r=1}^{n}r^3 - 6\sum_{r=1}^{n}r^2 - 2\sum_{r=1}^{n}r$ | M1 | Splitting the sum into three separate sums. PI by m1 line below.
$= 4\times\frac{1}{4}n^2(n+1)^2 - 6\times\frac{1}{6}n(n+1)(2n+1) - 2\times\frac{1}{2}n(n+1)$ | m1 | Substitution of the three summations from FB into $a\sum r^3 + b\sum r^2 + c\sum r$
$= n^2(n+1)^2 - n(n+1)(2n+1) - n(n+1)$ | A1 | PI by later expressions
$= n(n+1)[n(n+1)-(2n+1)-1]$ | m1 | Taking out factor $n(n+1)$ from correct expressions
$= n(n+1)[n^2-n-2]$ | A1 |
$= n(n+1)(n+1)(n-2)\ \left(= n(n-2)(n+1)^2\ (p=-2, q=1)\right)$ | A1 |
### Part (b):
$\displaystyle\sum_{r=11}^{20} 2r(2r^2-3r-1) = \sum_{r=1}^{20} 2r(2r^2-3r-1) - \sum_{r=1}^{10} 2r(2r^2-3r-1)$ | M1 | $\sum_{r=1}^{20}\cdots - \sum_{r=1}^{10}\cdots$. PI by next line (ft c's $p$ & $q$)
$= 20(20+p)(20+q)^2 - 10(10+p)(10+q)^2$
$= 20\times18\times21^2 - 10\times8\times11^2 = 158760 - 9680 = 149080$ | A1 | NMS 0/2. A0 if not showing use of fully factorised form.
---8
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\sum _ { r = 1 } ^ { n } 2 r \left( 2 r ^ { 2 } - 3 r - 1 \right) = n ( n + p ) ( n + q ) ^ { 2 }$$
where $p$ and $q$ are integers to be found.
\item Hence find the value of
$$\sum _ { r = 11 } ^ { 20 } 2 r \left( 2 r ^ { 2 } - 3 r - 1 \right)$$
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2013 Q8 [8]}}