| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 7 |
| 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 +0.3 Part (a) is a routine algebraic manipulation using standard summation formulae (splitting the sum, factoring, and simplifying to match the given form). Part (b) requires the standard technique of subtracting cumulative sums. While this is Further Maths content, it's a textbook exercise requiring only mechanical application of known methods with no novel insight. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Mark |
| (a) | \(\sum r^2(4r - 3) = 4\sum r^3 - 3\sum r^2\) ... | M1 |
| \(= 4\left(\frac{1}{4}n\right)^2(n+1)^2 - 3\left(\frac{1}{6}\right)n(n+1)(2n+1)\) | m1 | Substitution of the two summations from FB |
| \(= n(n+1)\left[n(n+1) - \frac{1}{2}(2n+1)\right]\) | m1 | Taking out common factors n and n+1 |
| Sum = \(\frac{1}{2}n(n+1)(2n^2 - 1)\) | A1 | 5 marks; Remaining expression eg our [...] in ACF not just simplified to AG; Be convinced as form of answer is given, penalise any jumps or backward steps |
| (b) | \(\sum_{r=20}^{40} r^2(4r-3)\) | M1 |
| \(= \sum_{r=1}^{40} r^2(4r-3) - \sum_{r=1}^{19} r^2(4r-3)\) | ||
| \(= 20(41)(3199) - 9.5(20)(721) = 2623180 - 136990\) | ||
| \(\sum_{r=20}^{40} r^2(4r-3) = 2486190\) | A1 | 2 marks; 2486190; Since 'Hence' NMS 0/2; SC \(\sum_{r=1}^{40} ... - \sum_{r=1}^{20} ...\) clearly attempted and evaluated to 2455390 scores B1 |
| Part | Answer/Working | Mark | Guidance |
|------|---|---|---|
| (a) | $\sum r^2(4r - 3) = 4\sum r^3 - 3\sum r^2$ ... | M1 | Splitting up the sum into two separate sums. PI by next line |
| | $= 4\left(\frac{1}{4}n\right)^2(n+1)^2 - 3\left(\frac{1}{6}\right)n(n+1)(2n+1)$ | m1 | Substitution of the two summations from FB |
| | $= n(n+1)\left[n(n+1) - \frac{1}{2}(2n+1)\right]$ | m1 | Taking out common factors n and n+1 |
| | Sum = $\frac{1}{2}n(n+1)(2n^2 - 1)$ | A1 | 5 marks; Remaining expression eg our [...] in ACF not just simplified to AG; Be convinced as form of answer is given, penalise any jumps or backward steps |
| (b) | $\sum_{r=20}^{40} r^2(4r-3)$ | M1 | Attempt to take S(19) from S(40) using part (a) |
| | $= \sum_{r=1}^{40} r^2(4r-3) - \sum_{r=1}^{19} r^2(4r-3)$ | | |
| | $= 20(41)(3199) - 9.5(20)(721) = 2623180 - 136990$ | | |
| | $\sum_{r=20}^{40} r^2(4r-3) = 2486190$ | A1 | 2 marks; 2486190; Since 'Hence' NMS 0/2; SC $\sum_{r=1}^{40} ... - \sum_{r=1}^{20} ...$ clearly attempted and evaluated to 2455390 scores B1 |4
\begin{enumerate}[label=(\alph*)]
\item Use the formulae for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ to show that
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 4 r - 3 ) = k n ( n + 1 ) \left( 2 n ^ { 2 } - 1 \right)$$
where $k$ is a constant.
\item Hence evaluate
$$\sum _ { r = 20 } ^ { 40 } r ^ { 2 } ( 4 r - 3 )$$
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2012 Q4 [7]}}