| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear transformations |
| Type | Extract enlargement and rotation parameters |
| Difficulty | Standard +0.3 This is a standard Core Pure AS question on matrix transformations requiring routine techniques: computing determinant (non-singular check), using |det M| for area scaling, and extracting enlargement/rotation parameters from a 2×2 matrix. While multi-part, each step follows textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(n=1\): \(\sum_{r=1}^{1} r^2 = 1\) and \(\frac{1}{6}n(n+1)(2n+1) = \frac{1}{6}(1)(2)(3) = 1\) | B1 | Checks \(n=1\) works for both sides of the general statement |
| Assume general statement true for \(n=k\); so assume \(\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k+1)(2k+1)\) is true | M1 | Assumes (general result) true for \(n=k\) |
| \(\sum_{r=1}^{k+1} r^2 = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2\) | M1 | Attempts to add \((k+1)\)th term to the sum of \(k\) terms |
| \(= \frac{1}{6}(k+1)(2k^2+7k+6)\) | A1 | Correct algebraic work leading to either \(\frac{1}{6}(k+1)(2k^2+7k+6)\) or \(\frac{1}{6}(k+2)(2k^2+5k+3)\) or \(\frac{1}{6}(2k+3)(k^2+3k+2)\) |
| \(= \frac{1}{6}(k+1)(k+2)(2k+3) = \frac{1}{6}(k+1)(\{k+1\}+1)(2\{k+1\}+1)\) | A1 | Correct algebraic work leading to \(\frac{1}{6}(k+1)(\{k+1\}+1)(2\{k+1\}+1)\) |
| Then general result is true for \(n=k+1\). As the general result has been shown to be true for \(n=1\), then the general result is true for all \(n \in \mathbb{Z}^+\) | A1 | cso leading to a correct induction statement conveying all three underlined points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=1}^{n} r(r+6)(r-6) = \sum_{r=1}^{n}(r^3 - 36r)\) | — | Expansion |
| \(= \frac{1}{4}n^2(n+1)^2 - \frac{36}{2}n(n+1)\) | M1, A1 | Substitutes at least one standard formula; correct expression |
| \(= \frac{1}{4}n(n+1)\big[n(n+1) - 72\big]\) | M1 | Depends on previous M mark; attempt to factorise at least \(n(n+1)\) |
| \(= \frac{1}{4}n(n+1)(n-8)(n+9)\) * cso | A1* | Obtains \(\frac{1}{4}n(n+1)(n-8)(n+9)\) by cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{4}n(n+1)(n-8)(n+9) = \frac{17}{6}n(n+1)(2n+1)\) | M1 | Sets their part (b) answer equal to \(\frac{17}{6}n(n+1)(2n+1)\) |
| \(\frac{1}{4}(n-8)(n+9) = \frac{17}{6}(2n+1)\) | M1 | Cancels out \(n(n+1)\) from both sides |
| \(3n^2 - 65n - 250 = 0\) | A1 | — |
| \((3n+10)(n-25) = 0\) | M1 | Valid method for solving a 3-term quadratic |
| As \(n\) must be a positive integer, \(n = 25\) | A1 | Only one solution of \(n=25\) |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: $\sum_{r=1}^{1} r^2 = 1$ and $\frac{1}{6}n(n+1)(2n+1) = \frac{1}{6}(1)(2)(3) = 1$ | B1 | Checks $n=1$ works for both sides of the general statement |
| Assume general statement true for $n=k$; so assume $\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k+1)(2k+1)$ is true | M1 | Assumes (general result) true for $n=k$ |
| $\sum_{r=1}^{k+1} r^2 = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2$ | M1 | Attempts to add $(k+1)$th term to the sum of $k$ terms |
| $= \frac{1}{6}(k+1)(2k^2+7k+6)$ | A1 | Correct algebraic work leading to either $\frac{1}{6}(k+1)(2k^2+7k+6)$ or $\frac{1}{6}(k+2)(2k^2+5k+3)$ or $\frac{1}{6}(2k+3)(k^2+3k+2)$ |
| $= \frac{1}{6}(k+1)(k+2)(2k+3) = \frac{1}{6}(k+1)(\{k+1\}+1)(2\{k+1\}+1)$ | A1 | Correct algebraic work leading to $\frac{1}{6}(k+1)(\{k+1\}+1)(2\{k+1\}+1)$ |
| Then general result is true for $n=k+1$. As the general result has been shown to be true for $n=1$, then the general result is true for all $n \in \mathbb{Z}^+$ | A1 | cso leading to a correct induction statement conveying all three underlined points |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{n} r(r+6)(r-6) = \sum_{r=1}^{n}(r^3 - 36r)$ | — | Expansion |
| $= \frac{1}{4}n^2(n+1)^2 - \frac{36}{2}n(n+1)$ | M1, A1 | Substitutes at least one standard formula; correct expression |
| $= \frac{1}{4}n(n+1)\big[n(n+1) - 72\big]$ | M1 | Depends on previous M mark; attempt to factorise at least $n(n+1)$ |
| $= \frac{1}{4}n(n+1)(n-8)(n+9)$ * cso | A1* | Obtains $\frac{1}{4}n(n+1)(n-8)(n+9)$ by cso |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{4}n(n+1)(n-8)(n+9) = \frac{17}{6}n(n+1)(2n+1)$ | M1 | Sets their part (b) answer equal to $\frac{17}{6}n(n+1)(2n+1)$ |
| $\frac{1}{4}(n-8)(n+9) = \frac{17}{6}(2n+1)$ | M1 | Cancels out $n(n+1)$ from both sides |
| $3n^2 - 65n - 250 = 0$ | A1 | — |
| $(3n+10)(n-25) = 0$ | M1 | Valid method for solving a 3-term quadratic |
| As $n$ must be a positive integer, $n = 25$ | A1 | Only one solution of $n=25$ |
---\begin{enumerate}
\item (a) Prove by induction that for all positive integers $n$,
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$
(b) Use the standard results for $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that for all positive integers $n$,
$$\sum _ { r = 1 } ^ { n } r ( r + 6 ) ( r - 6 ) = \frac { 1 } { 4 } n ( n + 1 ) ( n - 8 ) ( n + 9 )$$
(c) Hence find the value of $n$ that satisfies
$$\sum _ { r = 1 } ^ { n } r ( r + 6 ) ( r - 6 ) = 17 \sum _ { r = 1 } ^ { n } r ^ { 2 }$$
\hfill \mbox{\textit{Edexcel CP AS Q6 [15]}}