| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Prove sum formula |
| Difficulty | Standard +0.3 Part (a) is a standard proof of the geometric series formula that appears in most C2 textbooks and requires straightforward algebraic manipulation. Part (b) is a direct application requiring identification of a=10, r=2, n=12 and substitution into the formula. Both parts are routine exercises with clear methods, making this slightly easier than average for A-level. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.04g Sigma notation: for sums of series1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(S_n = a + ar + ar^2 + \ldots + ar^{n-1}\) | B1 | |
| \(rS_n = ar + ar^2 + ar^3 + \ldots + ar^n\) | M1 | |
| Subtracting, \(S_n - rS_n = a - ar^n\) | M1 | |
| \((1-r)S_n = a(1-r^n)\) | M1 | |
| \(S_n = \frac{a(1-r^n)}{1-r}\) | A1 | |
| (b) GP: \(a = 10, r = 2\) | B2 | |
| \(S_{12} = \frac{10(2^{12}-1)}{2-1}\) | M1 A1 | |
| \(= 40950\) | A1 | (9) |
**(a)** $S_n = a + ar + ar^2 + \ldots + ar^{n-1}$ | B1 |
$rS_n = ar + ar^2 + ar^3 + \ldots + ar^n$ | M1 |
Subtracting, $S_n - rS_n = a - ar^n$ | M1 |
$(1-r)S_n = a(1-r^n)$ | M1 |
$S_n = \frac{a(1-r^n)}{1-r}$ | A1 |
**(b)** GP: $a = 10, r = 2$ | B2 |
$S_{12} = \frac{10(2^{12}-1)}{2-1}$ | M1 A1 |
$= 40950$ | A1 | (9)\begin{enumerate}[label=(\alph*)]
\item Prove that the sum of the first $n$ terms of a geometric series with first term $a$ and common ratio $r$ is given by
$$\frac{a(1-r^n)}{1-r}.$$ [4]
\item Evaluate $\sum_{r=1}^{12} (5 \times 2^r)$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q7 [9]}}