| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: find parameter from given term |
| Difficulty | Moderate -0.8 This is a straightforward recurrence relation question requiring only systematic substitution and basic algebra. Part (a) involves direct application of the formula three times, part (b) is solving a linear equation, and part (c) is simple addition of four terms. No problem-solving insight needed, just careful arithmetic—easier than average A-level questions. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u_2 = 3k-12\), \(u_3 = 3(u_2)-12\) | M1 | Attempt to use formula twice to find \(u_2\) and \(u_3\) |
| \(u_2 = 3k-12\), \(u_3 = 9k-48\) | A1 | Two correct simplified answers |
| \(u_4 = 3(9k-48)-12 = 27k-156\) (ft their \(u_3\)) | M1, A1ft | Attempt again to find \(u_4\); 4th term correct and simplified, follow through their \(u_3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(27k - 156 = 15\) so \(k =\) | M1 | Put their 4th term (not 5th) equal to 15 and attempt to find \(k\) |
| \(k = 6\frac{1}{3}\) or \(\frac{19}{3}\) or \(6.33\) (3sf) | A1 | Accept any correct fraction or decimal answer (allow 6.33 or better) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{i=1}^{4} u_i = 6\frac{1}{3} + 7 + 9 + 15\) or \(\sum_{i=1}^{4} u_i = k + 3k-12 + 9k-48 + 27k-156\) | M1 | Uses 1st term and their following 3 terms with plus signs (either numerical or in terms of \(k\)); must use iteration terms not AP/GP formula |
| \(= 40k - 216 = 37\frac{1}{3}\) or \(\frac{112}{3}\) | A1ft, A1cao | A1ft for \(40k-216\) or follow through on their \(k\); A1 obtains \(37\frac{1}{3}\) (must be exact) |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u_2 = 3k-12$, $u_3 = 3(u_2)-12$ | M1 | Attempt to use formula twice to find $u_2$ and $u_3$ |
| $u_2 = 3k-12$, $u_3 = 9k-48$ | A1 | Two correct simplified answers |
| $u_4 = 3(9k-48)-12 = 27k-156$ (ft their $u_3$) | M1, A1ft | Attempt again to find $u_4$; 4th term correct and simplified, follow through their $u_3$ |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $27k - 156 = 15$ so $k =$ | M1 | Put their 4th term (not 5th) equal to 15 and attempt to find $k$ |
| $k = 6\frac{1}{3}$ or $\frac{19}{3}$ or $6.33$ (3sf) | A1 | Accept any correct fraction or decimal answer (allow 6.33 or better) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{i=1}^{4} u_i = 6\frac{1}{3} + 7 + 9 + 15$ or $\sum_{i=1}^{4} u_i = k + 3k-12 + 9k-48 + 27k-156$ | M1 | Uses 1st term and their following 3 terms **with plus signs** (either numerical or in terms of $k$); must use iteration terms not AP/GP formula |
| $= 40k - 216 = 37\frac{1}{3}$ or $\frac{112}{3}$ | A1ft, A1cao | A1ft for $40k-216$ or follow through on their $k$; A1 obtains $37\frac{1}{3}$ (must be exact) |\begin{enumerate}
\item A sequence is defined by
\end{enumerate}
$$\begin{aligned}
u _ { 1 } & = k \\
u _ { n + 1 } & = 3 u _ { n } - 12 , \quad n \geqslant 1
\end{aligned}$$
where $k$ is a constant.\\
(a) Write down fully simplified expressions for $u _ { 2 } , u _ { 3 }$ and $u _ { 4 }$ in terms of $k$.
Given that $u _ { 4 } = 15$\\
(b) find the value of $k$,\\
(c) find $\sum _ { i = 1 } ^ { 4 } u _ { i }$, giving an exact numerical answer.\\
\hfill \mbox{\textit{Edexcel C12 2015 Q8 [9]}}