| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Standard summation formula application |
| Difficulty | Standard +0.3 This is a structured multi-part summation question requiring expansion of r²(r+1), application of standard summation formulas, algebraic manipulation to find constants, and solving a polynomial equation. While it involves multiple steps and careful algebra, each part follows directly from the previous with clear guidance ('hence'), and the techniques are standard for Core Pure AS level. The question is slightly above average difficulty due to the length and algebraic manipulation required, but doesn't require novel insight or particularly challenging problem-solving. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n} r^2(r+1) = \sum r^3 + \sum r^2 = \frac{1}{4}n^2(n+1)^2 + \frac{1}{6}n(n+1)(2n+1)\) | M1, A1 | Substitutes at least one standard formula into expanded expression; both formulae correct |
| \(= \frac{1}{12}n(n+1)\left[3n(n+1)+2(2n+1)\right]\) | dM1 | Factorises correctly |
| \(= \frac{1}{12}n(n+1)\left[3n^2+7n+2\right] = \frac{1}{12}n(n+1)(n+2)(3n+1)\) cso | A1 | Correct factorised form, cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=k+1}^{3k} r^2(r+1) = \frac{1}{12}(3k)(3k+1)(3k+2)(9k+1) - \frac{1}{12}(k)(k+1)(k+2)(3k+1)\) | M1 | Correct strategy using result from (a) |
| \(= \frac{1}{12}k(3k+1)\left[3(3k+2)(9k+1)-(k+1)(k+2)\right]\) | M1 | Correct factorisation step |
| \(= \frac{1}{12}k(3k+1)\left[80k^2+60k+4\right] = \frac{1}{3}k(3k+1)(20k^2+15k+1)\) cso | A1 | Correct simplified result, cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{25}{3}k(3k+1)(20k^2+15k+1) = 192k^3(3k+1)\) leading to \(76k^2 - 375k - 25 = 0\) or \(76k^3-375k^2-25k=0\) or other valid polynomial equation | M1 | Sets up and simplifies to a polynomial equation |
| Solves their polynomial equation for \(k\) | M1 | Attempts solution of their polynomial |
| \(k = 5\) (only) | A1 | Correct unique positive integer solution |
# Question 3:
## Part (a)
| $\sum_{r=1}^{n} r^2(r+1) = \sum r^3 + \sum r^2 = \frac{1}{4}n^2(n+1)^2 + \frac{1}{6}n(n+1)(2n+1)$ | M1, A1 | Substitutes at least one standard formula into expanded expression; both formulae correct |
| $= \frac{1}{12}n(n+1)\left[3n(n+1)+2(2n+1)\right]$ | dM1 | Factorises correctly |
| $= \frac{1}{12}n(n+1)\left[3n^2+7n+2\right] = \frac{1}{12}n(n+1)(n+2)(3n+1)$ cso | A1 | Correct factorised form, cso |
## Part (b)
| $\sum_{r=k+1}^{3k} r^2(r+1) = \frac{1}{12}(3k)(3k+1)(3k+2)(9k+1) - \frac{1}{12}(k)(k+1)(k+2)(3k+1)$ | M1 | Correct strategy using result from (a) |
| $= \frac{1}{12}k(3k+1)\left[3(3k+2)(9k+1)-(k+1)(k+2)\right]$ | M1 | Correct factorisation step |
| $= \frac{1}{12}k(3k+1)\left[80k^2+60k+4\right] = \frac{1}{3}k(3k+1)(20k^2+15k+1)$ cso | A1 | Correct simplified result, cso |
## Part (c)
| $\frac{25}{3}k(3k+1)(20k^2+15k+1) = 192k^3(3k+1)$ leading to $76k^2 - 375k - 25 = 0$ or $76k^3-375k^2-25k=0$ or other valid polynomial equation | M1 | Sets up and simplifies to a polynomial equation |
| Solves their polynomial equation for $k$ | M1 | Attempts solution of their polynomial |
| $k = 5$ (only) | A1 | Correct unique positive integer solution |\begin{enumerate}
\item (a) Use the standard results for summations to show that, for all positive integers $n$,
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r + 1 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( a n + b )$$
where $a$ and $b$ are integers to be determined.\\
(b) Hence show that, for all positive integers $k$,
$$\sum _ { r = k + 1 } ^ { 3 k } r ^ { 2 } ( r + 1 ) = \frac { 1 } { 3 } k ( 3 k + 1 ) \left( A k ^ { 2 } + B k + C \right)$$
where $A , B$ and $C$ are integers to be determined.\\
(c) Hence, using algebra and making your method clear, determine the value of $k$ for which
$$25 \sum _ { r = k + 1 } ^ { 3 k } r ^ { 2 } ( r + 1 ) = 192 k ^ { 3 } ( 3 k + 1 )$$
\hfill \mbox{\textit{Edexcel CP AS 2024 Q3 [10]}}