| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Standard summation formula application |
| Difficulty | Standard +0.3 This is a multi-part question requiring standard summation formulae application, pattern recognition for a periodic tangent function, and algebraic manipulation. Part (a) is routine algebra with given formulae, part (b) requires spotting the period-3 pattern of tan(60r°), and part (c) combines these results with some careful index manipulation. While it has multiple steps, each component uses standard A-level techniques without requiring novel insight or particularly sophisticated problem-solving. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=14.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\sum_{r=1}^{n}(3r^2-17r-25) = 3\times\frac{n}{6}(n+1)(2n+1) - 17\times\frac{1}{2}n(n+1) - \ldots\) | M1 | Applies formulas for sum of integers and sum of squares |
| \(= 3\times\frac{n}{6}(n+1)(2n+1) - 17\times\frac{1}{2}n(n+1) - 25n\) | A1 | Correct unsimplified expression including \(25n\) |
| \(= n\left(\frac{1}{2}(2n^2+3n+1) - \frac{17}{2}(n+1) - 25\right)\) or \(= \frac{n}{2}\left((2n^2+3n+1)-17(n+1)-50\right)\) | M1 | Expands and factors out \(n\) or \(\frac{1}{2}n\) |
| \(= n(n^2-7n-33)\) (so \(A=7\) and \(B=33\)) | A1 cso | Correct proof, no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\sum_{r=1}^{3k} r\tan(60r)^\circ = \tan(60)^\circ + 2\tan(120)^\circ + 3\tan(180)^\circ + 4\tan(240)^\circ + 5\tan(300)^\circ + 6\tan(360)^\circ + \ldots\) \(= (\sqrt{3}-2\sqrt{3}+0)+(4\sqrt{3}-5\sqrt{3}+0)+\ldots\) | M1 | Writes out first few terms (at least 3) and identifies repeating pattern |
| Since tan has period \(180°\), \(\tan(60r)°\) repeats every three terms and each group of three terms results in \(-\sqrt{3}\) as a sum, so with \(k\) groups the sum is \(-k\sqrt{3}\) | A1 | Correct explanation identifying \(-\sqrt{3}\) is sum of each group of three terms |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\sum_{r=5}^{n}(3r^2-17r-25) = \sum_{r=1}^{n}(3r^2-17r-25) - \sum_{r=1}^{4}(3r^2-17r-25)\) | M1 | Applies formula from (a) as difference of two summations with either 4 or 5 as limit on second sum |
| \(= n(n^2-7n-33) - 4(4^2-7\times4-33)\) \(= n(n^2-7n-33)+180\) | A1 | Correct expression for left-hand side in terms of \(n\) |
| \(\sum_{r=6}^{3n} r\tan(60r)^\circ = -n\sqrt{3}+2\sqrt{3}\) (allow \(-n\sqrt{3}-{-2\sqrt{3}}\)) | B1 | Correct expression for right-hand side sum. Allow if lower limit 6 used instead of 5 |
| \(\Rightarrow n(n^2-7n-33)+180 = 15\left[-n\sqrt{3}+2\sqrt{3}\right]^2\) \(\Rightarrow n^3-7n^2-33n+180 = 15(3n^2-12n+12)\) \(\Rightarrow n^3-52n^2+147n = 0\) | M1 | Both sides expanded and terms gathered to reach simplified cubic in \(n\) |
| \(\Rightarrow n^3-52n^2+147n=0 \Rightarrow n = \ldots\) | M1 | Solves cubic equation (may use calculator, may divide by \(n\) and solve quadratic) |
| \(n = 49\) (reject \(n=3\) as summation on left undefined; accept \(n=0\) also given) | A1 | Selects correct value; \(n=3\) must be rejected, accept if \(n=0\) and \(n=49\) both given |
# Question 5:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}(3r^2-17r-25) = 3\times\frac{n}{6}(n+1)(2n+1) - 17\times\frac{1}{2}n(n+1) - \ldots$ | M1 | Applies formulas for sum of integers and sum of squares |
| $= 3\times\frac{n}{6}(n+1)(2n+1) - 17\times\frac{1}{2}n(n+1) - 25n$ | A1 | Correct unsimplified expression including $25n$ |
| $= n\left(\frac{1}{2}(2n^2+3n+1) - \frac{17}{2}(n+1) - 25\right)$ or $= \frac{n}{2}\left((2n^2+3n+1)-17(n+1)-50\right)$ | M1 | Expands and factors out $n$ or $\frac{1}{2}n$ |
| $= n(n^2-7n-33)$ (so $A=7$ and $B=33$) | A1 cso | Correct proof, no errors seen |
**(4 marks)**
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{3k} r\tan(60r)^\circ = \tan(60)^\circ + 2\tan(120)^\circ + 3\tan(180)^\circ + 4\tan(240)^\circ + 5\tan(300)^\circ + 6\tan(360)^\circ + \ldots$ $= (\sqrt{3}-2\sqrt{3}+0)+(4\sqrt{3}-5\sqrt{3}+0)+\ldots$ | M1 | Writes out first few terms (at least 3) and identifies repeating pattern |
| Since tan has period $180°$, $\tan(60r)°$ repeats every three terms and each group of three terms results in $-\sqrt{3}$ as a sum, so with $k$ groups the sum is $-k\sqrt{3}$ | A1 | Correct explanation identifying $-\sqrt{3}$ is sum of each group of three terms |
**(2 marks)**
## Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\sum_{r=5}^{n}(3r^2-17r-25) = \sum_{r=1}^{n}(3r^2-17r-25) - \sum_{r=1}^{4}(3r^2-17r-25)$ | M1 | Applies formula from (a) as difference of two summations with either 4 or 5 as limit on second sum |
| $= n(n^2-7n-33) - 4(4^2-7\times4-33)$ $= n(n^2-7n-33)+180$ | A1 | Correct expression for left-hand side in terms of $n$ |
| $\sum_{r=6}^{3n} r\tan(60r)^\circ = -n\sqrt{3}+2\sqrt{3}$ (allow $-n\sqrt{3}-{-2\sqrt{3}}$) | B1 | Correct expression for right-hand side sum. Allow if lower limit 6 used instead of 5 |
| $\Rightarrow n(n^2-7n-33)+180 = 15\left[-n\sqrt{3}+2\sqrt{3}\right]^2$ $\Rightarrow n^3-7n^2-33n+180 = 15(3n^2-12n+12)$ $\Rightarrow n^3-52n^2+147n = 0$ | M1 | Both sides expanded and terms gathered to reach simplified cubic in $n$ |
| $\Rightarrow n^3-52n^2+147n=0 \Rightarrow n = \ldots$ | M1 | Solves cubic equation (may use calculator, may divide by $n$ and solve quadratic) |
| $n = 49$ (reject $n=3$ as summation on left undefined; accept $n=0$ also given) | A1 | Selects correct value; $n=3$ must be rejected, accept if $n=0$ and $n=49$ both given |
**(6 marks)**\begin{enumerate}
\item (a) Use the standard summation formulae to show that, for $n \in \mathbb { N }$,
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 2 } - 17 r - 25 \right) = n \left( n ^ { 2 } - A n - B \right)$$
where $A$ and $B$ are integers to be determined.\\
(b) Explain why, for $k \in \mathbb { N }$,
$$\sum _ { r = 1 } ^ { 3 k } r \tan ( 60 r ) ^ { \circ } = - k \sqrt { 3 }$$
Using the results from part (a) and part (b) and showing all your working,\\
(c) determine any value of $n$ that satisfies
$$\sum _ { r = 5 } ^ { n } \left( 3 r ^ { 2 } - 17 r - 25 \right) = 15 \left[ \sum _ { r = 6 } ^ { 3 n } r \tan ( 60 r ) ^ { \circ } \right] ^ { 2 }$$
\hfill \mbox{\textit{Edexcel CP AS 2022 Q5 [12]}}