| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Standard summation formula application |
| Difficulty | Moderate -0.3 This is a straightforward application of standard summation formulas. Part (a) requires expanding (5r-2)², applying Σr² and Σr formulas, then algebraic simplification to match the given form. Part (b) involves solving a quadratic equation. Both parts are routine A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \((5r-2)^2 = 25r^2 - 20r + 4\) | B1 | 1.1b – Correct expansion |
| \(\sum_{r=1}^{n} 25r^2 - 20r + 4 = \frac{25}{6}n(n+1)(2n+1) - \frac{20}{2}n(n+1) + \ldots\) | M1 | 2.1 – Substitutes at least one standard formula |
| \(= \frac{25}{6}n(n+1)(2n+1) - \frac{20}{2}n(n+1) + 4n\) | A1 | 1.1b – Fully correct expression |
| \(= \frac{1}{6}n\left[25(2n^2+3n+1) - 60(n+1) + 24\right]\) | dM1 | 1.1b – Attempts to factorise \(\frac{1}{6}n\), dependent on M1 |
| \(= \frac{1}{6}n\left[50n^2 + 15n - 11\right]\) | A1 | 1.1b – Correct expression or correct values of \(a\), \(b\), \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\frac{1}{6}k\left[50k^2 + 15k - 11\right] = 94k^2\) | M1 | 1.1b – Uses part (a) result set equal to \(94k^2\), expands and collects |
| \(50k^3 - 549k^2 - 11k = 0\) or \(50k^2 - 549k - 11 = 0\) | A1 | 1.1b – Correct cubic or quadratic |
| \((k-11)(50k+1) = 0 \Rightarrow k = \ldots\) | M1 | 1.1b – Attempts to solve |
| \(k = 11\) (only) | A1 | 2.3 – Identifies correct value with no other values offered |
# Question 3:
## Part (a):
| Working | Mark | Notes |
|---------|------|-------|
| $(5r-2)^2 = 25r^2 - 20r + 4$ | B1 | 1.1b – Correct expansion |
| $\sum_{r=1}^{n} 25r^2 - 20r + 4 = \frac{25}{6}n(n+1)(2n+1) - \frac{20}{2}n(n+1) + \ldots$ | M1 | 2.1 – Substitutes at least one standard formula |
| $= \frac{25}{6}n(n+1)(2n+1) - \frac{20}{2}n(n+1) + 4n$ | A1 | 1.1b – Fully correct expression |
| $= \frac{1}{6}n\left[25(2n^2+3n+1) - 60(n+1) + 24\right]$ | dM1 | 1.1b – Attempts to factorise $\frac{1}{6}n$, dependent on M1 |
| $= \frac{1}{6}n\left[50n^2 + 15n - 11\right]$ | A1 | 1.1b – Correct expression or correct values of $a$, $b$, $c$ |
## Part (b):
| Working | Mark | Notes |
|---------|------|-------|
| $\frac{1}{6}k\left[50k^2 + 15k - 11\right] = 94k^2$ | M1 | 1.1b – Uses part (a) result set equal to $94k^2$, expands and collects |
| $50k^3 - 549k^2 - 11k = 0$ or $50k^2 - 549k - 11 = 0$ | A1 | 1.1b – Correct cubic or quadratic |
| $(k-11)(50k+1) = 0 \Rightarrow k = \ldots$ | M1 | 1.1b – Attempts to solve |
| $k = 11$ (only) | A1 | 2.3 – Identifies correct value with no other values offered |
**Total: 9 marks**
---\begin{enumerate}
\item (a) Use the standard results for summations to show that for all positive integers $n$
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } ( 5 r - 2 ) ^ { 2 } = \frac { 1 } { 6 } n \left( a n ^ { 2 } + b n + c \right)$$
where $a$, $b$ and $c$ are integers to be determined.\\
(b) Hence determine the value of $k$ for which
$$\sum _ { r = 1 } ^ { k } ( 5 r - 2 ) ^ { 2 } = 94 k ^ { 2 }$$
\hfill \mbox{\textit{Edexcel CP AS 2021 Q3 [9]}}