| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Sum of first n terms |
| Difficulty | Moderate -0.8 This is a straightforward geometric sequence question requiring only direct substitution into formulas and basic algebraic manipulation. Part (a) is simple evaluation, (b) is immediate from the formula, and (c) applies the standard sum formula with routine simplification. All steps are mechanical with no problem-solving or insight required, making it easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2(a) | \(u_1 = 12\) | B1 |
| \(u_2 = 3 \times 4^2 = 48\) | B1 | 2 marks |
| 2(b) | \(r = 4\) | B1 |
| 2(c)(i) | \(\{S_{12} = \frac{a(1-r^{12})}{1-r}\) | M1 |
| \(= \frac{12(1-4^{12})}{1-4}\) | A1ft | |
| \(= \frac{12(1-4^{12})}{-3} = -4(1-4^{12}) = 4^{13} - 4\) | A1 | 3 marks |
| 2(c)(ii) | \(\sum_{n=2}^{12} u_n = (4^{13} - 4) - u_1\) | B1 |
| \(= 67108848\) | B1 | 1 mark |
**2(a)** | $u_1 = 12$ | B1 | |
| | $u_2 = 3 \times 4^2 = 48$ | B1 | 2 marks | CSO AG (be convinced) |
**2(b)** | $r = 4$ | B1 | 1 mark |
**2(c)(i)** | $\{S_{12} = \frac{a(1-r^{12})}{1-r}$ | M1 | | OE Using a correct formula with $n = 12$ |
| | $= \frac{12(1-4^{12})}{1-4}$ | A1ft | | Ft on answer for $u_1$ in (a) and $r$ in (b) |
| | $= \frac{12(1-4^{12})}{-3} = -4(1-4^{12}) = 4^{13} - 4$ | A1 | 3 marks | CAO Accept $k = 13$ for $4^{13}$ term |
**2(c)(ii)** | $\sum_{n=2}^{12} u_n = (4^{13} - 4) - u_1$ | B1 | |
| | $= 67108848$ | B1 | 1 mark |
---2 The $n$th term of a geometric sequence is $u _ { n }$, where
$$u _ { n } = 3 \times 4 ^ { n }$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $u _ { 1 }$ and show that $u _ { 2 } = 48$.
\item Write down the common ratio of the geometric sequence.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the sum of the first 12 terms of the geometric sequence is $4 ^ { k } - 4$, where $k$ is an integer.
\item Hence find the value of $\sum _ { n = 2 } ^ { 12 } u _ { n }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2007 Q2 [7]}}