| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Standard summation formulae application |
| Difficulty | Standard +0.8 This question requires manipulating summation formulae to derive a non-standard result (sum of cubes of odd numbers), then solving a quartic equation by substitution. Part (a) needs algebraic manipulation of the given summation formulae, while part (b) involves setting up and solving n²(2n²-1) - (n-10)²(2(n-10)²-1) = 99800, which is more challenging than typical summation exercises but still follows a clear method once the setup is understood. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete attempt to find \(\sum(2r-1)^3\) or \(\sum(2r+1)^3\) by expanding and using three standard summation formulae | M1 | 3.1a |
| \(\sum_{r=1}^{n}(2r-1)^3 = \sum_{r=1}^{n}(8r^3-12r^2+6r-1) = 8\sum r^3 - 12\sum r^2 + 6\sum r - \sum 1\) | M1 | 1.1b |
| \(= 8\frac{n^2}{4}(n+1)^2 - 12\frac{n}{6}(n+1)(2n+1) + 6\frac{n}{2}(n+1) - n\) | M1, A1 | Applies result for at least three summations; correct unsimplified expression |
| Leading to \(= n^2(2n^2-1)\) cso | A1* | Multiplies out to achieve correct intermediate line clearly leading to result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=n}^{n+9}(2r-1)^3 = \sum_{r=1}^{n+9}(2r-1)^3 - \sum_{r=1}^{n-1}(2r-1)^3 = (n+9)^2(2(n+9)^2-1)-(n-1)^2(2(n-1)^2-1) = 99800\) | M1 | Uses part (a) to find sum of 10 consecutive odd cubes set equal to 99800 |
| \(80n^3 + 960n^2 + 5820n - 86760 = 0\) (or equivalent) | A1 | Correct simplified cubic equation |
| Solves cubic equation | dM1 | Uses calculator; dependent on previous M1 |
| \(n=6\), smallest number \(= 11\) | A1 | cao |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete attempt to find $\sum(2r-1)^3$ or $\sum(2r+1)^3$ by expanding and using three standard summation formulae | M1 | 3.1a |
| $\sum_{r=1}^{n}(2r-1)^3 = \sum_{r=1}^{n}(8r^3-12r^2+6r-1) = 8\sum r^3 - 12\sum r^2 + 6\sum r - \sum 1$ | M1 | 1.1b |
| $= 8\frac{n^2}{4}(n+1)^2 - 12\frac{n}{6}(n+1)(2n+1) + 6\frac{n}{2}(n+1) - n$ | M1, A1 | Applies result for at least three summations; correct unsimplified expression |
| Leading to $= n^2(2n^2-1)$ cso | A1* | Multiplies out to achieve correct intermediate line clearly leading to result |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=n}^{n+9}(2r-1)^3 = \sum_{r=1}^{n+9}(2r-1)^3 - \sum_{r=1}^{n-1}(2r-1)^3 = (n+9)^2(2(n+9)^2-1)-(n-1)^2(2(n-1)^2-1) = 99800$ | M1 | Uses part (a) to find sum of 10 consecutive odd cubes set equal to 99800 |
| $80n^3 + 960n^2 + 5820n - 86760 = 0$ (or equivalent) | A1 | Correct simplified cubic equation |
| Solves cubic equation | dM1 | Uses calculator; dependent on previous M1 |
| $n=6$, smallest number $= 11$ | A1 | cao |\begin{enumerate}
\item In this question you may assume the results for
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } r ^ { 3 } , \sum _ { r = 1 } ^ { n } r ^ { 2 } \text { and } \sum _ { r = 1 } ^ { n } r$$
(a) Show that the sum of the cubes of the first $n$ positive odd numbers is
$$n ^ { 2 } \left( 2 n ^ { 2 } - 1 \right)$$
The sum of the cubes of 10 consecutive positive odd numbers is 99800\\
(b) Use the answer to part (a) to determine the smallest of these 10 consecutive positive odd numbers.
\hfill \mbox{\textit{Edexcel CP2 2021 Q4 [9]}}