| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Standard summation formula application |
| Difficulty | Moderate -0.3 Part (a) is a routine algebraic manipulation using standard summation formulas (expanding (2r-1)², applying given formulas, simplifying to find a=4/3). Part (b) applies this result with straightforward identification of n (first odd 3-digit is 101=2(51)-1, last is 999=2(500)-1, so n=450). This is a standard textbook exercise requiring formula recall and careful arithmetic but no problem-solving insight or novel techniques. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2r-1)^2 = 4r^2 - 4r + 1\) | B1 | Correct expanded expression |
| \(\sum_{r=1}^{n}(2r-1)^2 = 4\sum_{r=1}^{n}r^2 - 4\sum_{r=1}^{n}r + \sum_{r=1}^{n}1\) | M1 | Substitutes at least one standard formula into expanded expression |
| \(= 4\cdot\frac{n}{6}(n+1)(2n+1) - 4\cdot\frac{n}{2}(n+1) + n\) | A1 | Fully correct unsimplified expression |
| \(= \frac{n}{3}\left[2(n+1)(2n+1) - 6(n+1) + 3\right]\) or \(= n\left[\frac{2}{3}(n+1)(2n+1) - 2(n+1) + 1\right]\) | dM1 | Dependent on previous M1. Attempts to factorise out \(n\). Must have \(n\) in every term. Condone a slip with one term as long as intention is clear |
| \(= \frac{n}{3}(4n^2 - 1)\) cso | A1 | Achieves correct answer with a correct intermediate line of working. Note: if uses \(\sum 1 = 1\) scores B1 M1 A0 M0 A0. An attempt at proof by induction may score B1 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=51}^{500}(2r-1)^2\) | B1 | Correct summation formula for sum of squares of all positive odd three-digit integers including limits. Can be implied by later work |
| \(\sum_{r=51}^{500}(2r-1)^2 = \sum_{r=1}^{500}(2r-1)^2 - \sum_{r=1}^{50}(2r-1)^2 = \frac{500}{3}(4(500)^2-1) - \frac{50}{3}(4(50)^2-1)\) | M1 | Uses answer to part (a) with \(\sum_{r=p}^{q}(2r-1)^2 = \sum_{r=1}^{q}(2r-1)^2 - \sum_{r=1}^{p-1}(2r-1)^2\) where \(p\), \(q\) numerical and \(q>p\) |
| \(= 166666500 - 166650\) | ||
| \(166\,499\,850\) | A1 | Correct value. Note: correct answer only scores B1 M0 A0, must be evidence of using answer to (a) |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2r-1)^2 = 4r^2 - 4r + 1$ | B1 | Correct expanded expression |
| $\sum_{r=1}^{n}(2r-1)^2 = 4\sum_{r=1}^{n}r^2 - 4\sum_{r=1}^{n}r + \sum_{r=1}^{n}1$ | M1 | Substitutes at least one standard formula into expanded expression |
| $= 4\cdot\frac{n}{6}(n+1)(2n+1) - 4\cdot\frac{n}{2}(n+1) + n$ | A1 | Fully correct unsimplified expression |
| $= \frac{n}{3}\left[2(n+1)(2n+1) - 6(n+1) + 3\right]$ or $= n\left[\frac{2}{3}(n+1)(2n+1) - 2(n+1) + 1\right]$ | dM1 | Dependent on previous M1. Attempts to factorise out $n$. Must have $n$ in every term. Condone a slip with one term as long as intention is clear |
| $= \frac{n}{3}(4n^2 - 1)$ cso | A1 | Achieves correct answer with a correct intermediate line of working. Note: if uses $\sum 1 = 1$ scores B1 M1 A0 M0 A0. An attempt at proof by induction may score B1 only |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=51}^{500}(2r-1)^2$ | B1 | Correct summation formula for sum of squares of all positive odd three-digit integers including limits. Can be implied by later work |
| $\sum_{r=51}^{500}(2r-1)^2 = \sum_{r=1}^{500}(2r-1)^2 - \sum_{r=1}^{50}(2r-1)^2 = \frac{500}{3}(4(500)^2-1) - \frac{50}{3}(4(50)^2-1)$ | M1 | Uses answer to part (a) with $\sum_{r=p}^{q}(2r-1)^2 = \sum_{r=1}^{q}(2r-1)^2 - \sum_{r=1}^{p-1}(2r-1)^2$ where $p$, $q$ numerical and $q>p$ |
| $= 166666500 - 166650$ | | |
| $166\,499\,850$ | A1 | Correct value. Note: correct answer only scores B1 M0 A0, must be evidence of using answer to (a) |
---\begin{enumerate}
\item (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that, for all positive integers $n$,
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 } = \frac { n } { 3 } \left( a n ^ { 2 } - 1 \right)$$
where $a$ is a constant to be determined.\\
(b) Hence determine the sum of the squares of all positive odd three-digit integers.
\hfill \mbox{\textit{Edexcel CP AS 2023 Q8 [8]}}