| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: evaluate sum |
| Difficulty | Moderate -0.3 This is a straightforward recurrence relation question requiring systematic substitution and algebraic manipulation. Part (a) is direct substitution, part (b) involves simple summation of three terms, and part (c) uses telescoping series recognition (the sum collapses to a₁₀₁ - a₁). While it requires careful algebra, it's a standard C1 exercise with no novel problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a_2 = 5 - ka_1 = 5 - 4k\) | M1 A1 | M1: Uses the recurrence relation correctly at least once. May be implied by \(a_2 = 5 - 4k\) or by use of \(a_3 = 5 - k(\text{their } a_2)\). A1: Two correct expressions – need not be simplified but must be seen in (a). Allow \(a_2 = 5 - k4\) and \(a_3 = 5 - 5k + k^2 4\) |
| \(a_3 = 5 - ka_2 = 5 - k(5-4k)\) | isw if necessary for \(a_3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=1}^{3}(1) = 1+1+1\) | B1 | Finds \(1+1+1\) or \(3\) somewhere in their solution (may be implied by e.g. \(5 + 6 - 4k + 6 - 5k + 4k^2\)). Note \(5 + 6 - 4k + 6 - 5k + 4k^2\) would score B1 and the M1 below |
| \(\sum_{r=1}^{3} a_r = 4 + \text{"} 5-4k\text{"} + \text{"} 5-5k+4k^2\text{"}\) | M1 | Adds 4 to their \(a_2\) and their \(a_3\) where \(a_2\) and \(a_3\) are functions of \(k\) |
| \(\sum_{r=1}^{3}(1 + a_r) = 17 - 9k + 4k^2\) | A1 | cao but condone '\(= 0\)' after the expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(500\) | B1 | cao |
# Question 6:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a_2 = 5 - ka_1 = 5 - 4k$ | M1 A1 | M1: Uses the recurrence relation correctly at least once. May be implied by $a_2 = 5 - 4k$ or by use of $a_3 = 5 - k(\text{their } a_2)$. A1: Two correct expressions – need not be simplified but must be seen in (a). Allow $a_2 = 5 - k4$ and $a_3 = 5 - 5k + k^2 4$ |
| $a_3 = 5 - ka_2 = 5 - k(5-4k)$ | | isw if necessary for $a_3$ |
**[2]**
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{3}(1) = 1+1+1$ | B1 | Finds $1+1+1$ or $3$ somewhere in their solution (may be implied by e.g. $5 + 6 - 4k + 6 - 5k + 4k^2$). Note $5 + 6 - 4k + 6 - 5k + 4k^2$ would score B1 and the M1 below |
| $\sum_{r=1}^{3} a_r = 4 + \text{"} 5-4k\text{"} + \text{"} 5-5k+4k^2\text{"}$ | M1 | Adds 4 to their $a_2$ and their $a_3$ where $a_2$ and $a_3$ are functions of $k$ |
| $\sum_{r=1}^{3}(1 + a_r) = 17 - 9k + 4k^2$ | A1 | cao but condone '$= 0$' after the expression |
**Allow full marks in (b) for correct answer only**
**[3]**
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $500$ | B1 | cao |
**[1]**
**[6 marks total]**
---6. A sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is defined by
$$\begin{aligned}
a _ { 1 } & = 4 \\
a _ { n + 1 } & = 5 - k a _ { n } , \quad n \geqslant 1
\end{aligned}$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Write down expressions for $a _ { 2 }$ and $a _ { 3 }$ in terms of $k$.
Find
\item $\sum _ { r = 1 } ^ { 3 } \left( 1 + a _ { r } \right)$ in terms of $k$, giving your answer in its simplest form,
\item $\sum _ { r = 1 } ^ { 100 } \left( a _ { r + 1 } + k a _ { r } \right)$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2016 Q6 [6]}}