| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Alternating series summation |
| Difficulty | Standard +0.8 This is a multi-part question requiring manipulation of alternating series, algebraic expansion, and application of standard summation formulae. Part (a) requires understanding of series notation and parity arguments. Part (b) involves expanding $((-1)^r + 2r)^2$, separating into alternating and non-alternating components, and applying summation formulae with careful algebraic manipulation. Part (c) requires using the difference of two sums. While systematic, it demands careful bookkeeping across multiple steps and is more challenging than typical Core Pure 1 summation questions. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| All even terms positive, all odd terms negative, or \(\sum_{r=1}^{2n}(-1)^r f(r) = -f(1)+f(2)-f(3)+f(4)-...\) | M1 | AO2.4 - stating odd terms negative/even terms positive, or writing at least 4 terms |
| \(\sum_{r=1}^{2n}(-1)^r f(r) = f(2)+f(4)+...+f(2n) - (f(1)+f(3)+...+f(2n-1)) = \sum_{r=1}^{n}(f(2r)-f(2r-1))\) | A1* | AO3.1a - full convincing argument showing even terms grouped as \(f(2)+f(4)+...+f(2n)\) AND odd terms grouped as \(f(1)+f(3)+...+f(2n-1)\) AND conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=1}^{2n} r\left((-1)^r+2r\right)^2 = \sum_{r=1}^{2n} r\left(1+4r(-1)^r+4r^2\right)\) | M1 | AO2.1 - expands fully and applies \((-1)^{2r}=1\) or \(r(-1)^{2r}=r\) |
| \(= \frac{1}{2}(2n)(2n+1) + 4\frac{(2n)^2}{4}(2n+1)^2 + 4\sum_{r=1}^{2n}(-1)^r r^2\) | M1 | AO1.1b - applies correct formulae for sum of integers and sum of cubes to \(r\) and \(mr^3\) terms with \(2n\) |
| \(\sum_{r=1}^{2n}(-1)^r r^2 = \sum_{r=1}^{n}\left((2r)^2-(2r-1)^2\right)\) | M1 | AO3.1a - applies result of part (a) to the \((-1)^r\) term |
| \(4\sum_{r=1}^{2n}(-1)^r r^2 = 4\left(4\frac{n}{2}(n+1)-n\right)\) | B1 | AO1.1b - correct expression; depends on previous method mark |
| \(= n(2n+1) + 4(n(2n+1))^2 + 4n(2n+1)\) | dM1 | AO2.1 - takes out factor \(n(2n+1)\); depends on all previous M marks |
| \(= n(2n+1)\left(8n^2+4n+5\right)\) | A1* | AO1.1b - correct answer, no errors, suitable intermediate steps shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=14}^{50} r\left((-1)^r+2r\right)^2 = \sum_{r=1}^{50} r\left((-1)^r+2r\right)^2 - \sum_{r=1}^{13} r\left((-1)^r+2r\right)^2\) | M1 | AO1.1b - attempts to split sum to difference with upper limit 50 and lower limit 13 |
| \(= \sum_{r=1}^{50}(\cdots) - \sum_{r=1}^{12}(\cdots) - 13\times 25^2\) or \(= \sum_{r=1}^{50}(\cdots) - \sum_{r=1}^{14}(\cdots) + 14\times 29^2\) | M1 | AO3.1a - splits with even upper limits for both sums and evaluates balancing term correctly |
| \(= 25\times51\times5105 - 6\times13\times317 - 13\times25^2\) \((= 6508875 - 24726 - 8125)\) | M1 | AO2.1 - uses result from (b) at least once correctly with integer; \(n=25\) with upper limit 50, or \(n=6\) with upper limit 12 |
| \(= 6476024\) | A1 | AO1.1b |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| All even terms positive, all odd terms negative, or $\sum_{r=1}^{2n}(-1)^r f(r) = -f(1)+f(2)-f(3)+f(4)-...$ | M1 | AO2.4 - stating odd terms negative/even terms positive, or writing at least 4 terms |
| $\sum_{r=1}^{2n}(-1)^r f(r) = f(2)+f(4)+...+f(2n) - (f(1)+f(3)+...+f(2n-1)) = \sum_{r=1}^{n}(f(2r)-f(2r-1))$ | A1* | AO3.1a - full convincing argument showing even terms grouped as $f(2)+f(4)+...+f(2n)$ AND odd terms grouped as $f(1)+f(3)+...+f(2n-1)$ AND conclusion |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{2n} r\left((-1)^r+2r\right)^2 = \sum_{r=1}^{2n} r\left(1+4r(-1)^r+4r^2\right)$ | M1 | AO2.1 - expands fully and applies $(-1)^{2r}=1$ or $r(-1)^{2r}=r$ |
| $= \frac{1}{2}(2n)(2n+1) + 4\frac{(2n)^2}{4}(2n+1)^2 + 4\sum_{r=1}^{2n}(-1)^r r^2$ | M1 | AO1.1b - applies correct formulae for sum of integers and sum of cubes to $r$ and $mr^3$ terms with $2n$ |
| $\sum_{r=1}^{2n}(-1)^r r^2 = \sum_{r=1}^{n}\left((2r)^2-(2r-1)^2\right)$ | M1 | AO3.1a - applies result of part (a) to the $(-1)^r$ term |
| $4\sum_{r=1}^{2n}(-1)^r r^2 = 4\left(4\frac{n}{2}(n+1)-n\right)$ | B1 | AO1.1b - correct expression; depends on previous method mark |
| $= n(2n+1) + 4(n(2n+1))^2 + 4n(2n+1)$ | dM1 | AO2.1 - takes out factor $n(2n+1)$; depends on all previous M marks |
| $= n(2n+1)\left(8n^2+4n+5\right)$ | A1* | AO1.1b - correct answer, no errors, suitable intermediate steps shown |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=14}^{50} r\left((-1)^r+2r\right)^2 = \sum_{r=1}^{50} r\left((-1)^r+2r\right)^2 - \sum_{r=1}^{13} r\left((-1)^r+2r\right)^2$ | M1 | AO1.1b - attempts to split sum to difference with upper limit 50 and lower limit 13 |
| $= \sum_{r=1}^{50}(\cdots) - \sum_{r=1}^{12}(\cdots) - 13\times 25^2$ or $= \sum_{r=1}^{50}(\cdots) - \sum_{r=1}^{14}(\cdots) + 14\times 29^2$ | M1 | AO3.1a - splits with even upper limits for both sums and evaluates balancing term correctly |
| $= 25\times51\times5105 - 6\times13\times317 - 13\times25^2$ $(= 6508875 - 24726 - 8125)$ | M1 | AO2.1 - uses result from (b) at least once correctly with integer; $n=25$ with upper limit 50, or $n=6$ with upper limit 12 |
| $= 6476024$ | A1 | AO1.1b |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.\\
(a) Explain why, for $n \in \mathbb { N }$
\end{enumerate}
$$\sum _ { r = 1 } ^ { 2 n } ( - 1 ) ^ { r } \mathrm { f } ( r ) = \sum _ { r = 1 } ^ { n } ( \mathrm { f } ( 2 r ) - \mathrm { f } ( 2 r - 1 ) )$$
for any function $\mathrm { f } ( r )$.\\
(b) Use the standard summation formulae to show that, for $n \in \mathbb { N }$
$$\sum _ { r = 1 } ^ { 2 n } r \left( ( - 1 ) ^ { r } + 2 r \right) ^ { 2 } = n ( 2 n + 1 ) \left( 8 n ^ { 2 } + 4 n + 5 \right)$$
(c) Hence evaluate
$$\sum _ { r = 14 } ^ { 50 } r \left( ( - 1 ) ^ { r } + 2 r \right) ^ { 2 }$$
\hfill \mbox{\textit{Edexcel CP1 2023 Q7 [12]}}