| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: evaluate sum |
| Difficulty | Moderate -0.8 This is a straightforward recurrence relation question requiring only direct substitution and pattern recognition. Part (i) involves computing U₃ by plugging values into the formula, then recognizing the constant sequence to find the sum (20×4=80). Part (ii) is similar algebraic substitution with k. No problem-solving insight needed—purely mechanical application of the given formula, making it easier than average C1 questions. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(U_3=4\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{n=1}^{n=20} U_n = 4+4+4+\ldots+4\) or \(20\times4\) | M1 | Realising all 20 terms are 4 and sum required. Accept \(\frac{1}{2}\times20(2\times4+19\times0)\) or \(\frac{1}{2}\times20(4+4)\) with correct sum formula \(n=20\), \(a=4\), \(d=0\) or \(n=20\), \(a=4\), \(l=4\) |
| \(=80\) | A1 | cao. Correct answer with no working scores M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V_3=3k,\quad V_4=4k\) | B1, B1 | May score in (b) if clearly identified as \(V_3\) and \(V_4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{n=1}^{n=5}V_n=k+2k+3k+4k+5k=165\) or \(\frac{1}{2}\times5(2\times k+4\times k)=165\) or \(\frac{1}{2}\times5(k+5k)=165\) | M1 | Attempts \(V_5\), adds \(V_1,V_2,V_3,V_4,V_5\) and sets equal to 165. Or use of correct sum formula with \(a=k\), \(d=k\), \(n=5\) or \(a=k\), \(l=5k\), \(n=5\) AND sets equal to 165 |
| \(15k=165 \Rightarrow k=\ldots\) | M1 | Attempts to solve linear equation in \(k\) having set sum of first 5 terms equal to 165. Solving \(V_5=165\) scores no marks |
| \(k=11\) | A1 | cao and cso |
# Question 4:
## Part (i)(a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $U_3=4$ | B1 | cao |
## Part (i)(b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{n=1}^{n=20} U_n = 4+4+4+\ldots+4$ or $20\times4$ | M1 | Realising all 20 terms are 4 and sum required. Accept $\frac{1}{2}\times20(2\times4+19\times0)$ or $\frac{1}{2}\times20(4+4)$ with correct sum formula $n=20$, $a=4$, $d=0$ or $n=20$, $a=4$, $l=4$ |
| $=80$ | A1 | cao. Correct answer with no working scores M1A1 |
## Part (ii)(a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $V_3=3k,\quad V_4=4k$ | B1, B1 | May score in (b) if clearly identified as $V_3$ and $V_4$ |
## Part (ii)(b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{n=1}^{n=5}V_n=k+2k+3k+4k+5k=165$ **or** $\frac{1}{2}\times5(2\times k+4\times k)=165$ **or** $\frac{1}{2}\times5(k+5k)=165$ | M1 | Attempts $V_5$, adds $V_1,V_2,V_3,V_4,V_5$ and sets equal to 165. Or use of correct sum formula with $a=k$, $d=k$, $n=5$ or $a=k$, $l=5k$, $n=5$ AND sets equal to 165 |
| $15k=165 \Rightarrow k=\ldots$ | M1 | Attempts to solve linear equation in $k$ having **set sum of first 5 terms equal to 165**. Solving $V_5=165$ scores no marks |
| $k=11$ | A1 | cao and cso |(i) A sequence $U _ { 1 } , U _ { 2 } , U _ { 3 } , \ldots$ is defined by
$$\begin{gathered}
U _ { n + 2 } = 2 U _ { n + 1 } - U _ { n } , \quad n \geqslant 1 \\
U _ { 1 } = 4 \text { and } U _ { 2 } = 4
\end{gathered}$$
Find the value of
\begin{enumerate}[label=(\alph*)]
\item $U _ { 3 }$
\item $\sum _ { n = 1 } ^ { 20 } U _ { n }$\\
(ii) Another sequence $V _ { 1 } , V _ { 2 } , V _ { 3 } , \ldots$ is defined by\\
(a) Find $V _ { 3 }$ and $V _ { 4 }$ in terms of $k$.
$$\begin{gathered}
V _ { n + 2 } = 2 V _ { n + 1 } - V _ { n } , \quad n \geqslant 1 \\
V _ { 1 } = k \text { and } V _ { 2 } = 2 k , \text { where } k \text { is a constant }
\end{gathered}$$
a) Find $V _ { 3 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2015 Q4 [8]}}