| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Recursive sequence definition |
| Difficulty | Moderate -0.3 This is a straightforward geometric sequence question requiring identification of first term and common ratio (r=2/3), direct calculation of terms, application of the standard GP sum formula, and a simple limit argument. While multi-part with 4 sections, each part uses routine techniques with no novel insight required—slightly easier than average due to the standard nature of all components. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_2 = \frac{8}{3}\), \(u_3 = \frac{16}{9}\), \(u_4 = \frac{32}{27}\) | M1, A1 | M1: any one term is \(\frac{2}{3}\) the previous term; accept \(u_2 =\) awrt 2.67. A1: all 3 terms correct; accept exact equivalents \(2\frac{2}{3}\), \(1\frac{7}{9}\), \(1\frac{5}{27}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_{20} = 4 \times \left(\frac{2}{3}\right)^{19} = 0.00180\) or \(0.0018\) | M1, A1 cao | M1: uses correct nth term formula \(ar^{n-1}\) with \(a=4\), \(n=20\), \(r=\frac{2}{3}\) or awrt 0.7. Condone use of \(ar^{n-1}\) with \(a=\frac{8}{3}\), \(n=20\), \(r=\frac{2}{3}\). Expressions \(4\times\left(\frac{2}{3}\right)^{19}\), \(\frac{8}{3}\times\left(\frac{2}{3}\right)^{18}\), \(\frac{2^{21}}{3^{19}} \to \frac{2^{21}}{3^{19}}\) are correct. A1: accept \(0.0018\), \(0.00180\), \(1.80\times10^{-3}\) or \(1.8\times10^{-3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\sum_{i=1}^{16} u_i = \frac{4\left(1-\left(\frac{2}{3}\right)^{16}\right)}{1-\frac{2}{3}}\) | M1 | Uses correct sum formula \(S = \frac{a(r^n-1)}{r-1}\) or \(\frac{a(1-r^n)}{1-r}\) with \(a=4\), \(r=\frac{2}{3}\) or awrt 0.7, \(n=16\). Condone with \(a=\frac{8}{3}\), \(r=\frac{2}{3}\), \(n=16\) |
| Find \(12 - \text{their } \sum_{i=1}^{16} u_i\) | dM1 | Dependent on previous M mark |
| \(= 12 - 11.9817 =\) awrt \(0.0183\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 12 is the sum to infinity (and all terms are positive) so sum is less than 12 | B1 | Need a reason + minimal conclusion. E.g. "sum to infinity \(= 12\) and sum is less than 12". Or: \(\sum_{i=1}^{n} u_i = \frac{4(1-(\frac{2}{3})^n)}{1-\frac{2}{3}} = 12 - 12\left(\frac{2}{3}\right)^n\) and \(\left(\frac{2}{3}\right)^n > 0\) so is less than 12 |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_2 = \frac{8}{3}$, $u_3 = \frac{16}{9}$, $u_4 = \frac{32}{27}$ | M1, A1 | M1: any one term is $\frac{2}{3}$ the previous term; accept $u_2 =$ awrt 2.67. A1: all 3 terms correct; accept exact equivalents $2\frac{2}{3}$, $1\frac{7}{9}$, $1\frac{5}{27}$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_{20} = 4 \times \left(\frac{2}{3}\right)^{19} = 0.00180$ or $0.0018$ | M1, A1 cao | M1: uses correct nth term formula $ar^{n-1}$ with $a=4$, $n=20$, $r=\frac{2}{3}$ or awrt 0.7. Condone use of $ar^{n-1}$ with $a=\frac{8}{3}$, $n=20$, $r=\frac{2}{3}$. Expressions $4\times\left(\frac{2}{3}\right)^{19}$, $\frac{8}{3}\times\left(\frac{2}{3}\right)^{18}$, $\frac{2^{21}}{3^{19}} \to \frac{2^{21}}{3^{19}}$ are correct. A1: accept $0.0018$, $0.00180$, $1.80\times10^{-3}$ or $1.8\times10^{-3}$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\sum_{i=1}^{16} u_i = \frac{4\left(1-\left(\frac{2}{3}\right)^{16}\right)}{1-\frac{2}{3}}$ | M1 | Uses correct sum formula $S = \frac{a(r^n-1)}{r-1}$ or $\frac{a(1-r^n)}{1-r}$ with $a=4$, $r=\frac{2}{3}$ or awrt 0.7, $n=16$. Condone with $a=\frac{8}{3}$, $r=\frac{2}{3}$, $n=16$ |
| Find $12 - \text{their } \sum_{i=1}^{16} u_i$ | dM1 | Dependent on previous M mark |
| $= 12 - 11.9817 =$ awrt $0.0183$ | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 12 is the **sum to infinity** (and all terms are **positive**) so sum is less than 12 | B1 | Need a reason + minimal conclusion. E.g. "sum to infinity $= 12$ **and** sum is less than 12". **Or:** $\sum_{i=1}^{n} u_i = \frac{4(1-(\frac{2}{3})^n)}{1-\frac{2}{3}} = 12 - 12\left(\frac{2}{3}\right)^n$ and $\left(\frac{2}{3}\right)^n > 0$ so is less than 12 |
---10. A sequence is defined by
$$\begin{aligned}
u _ { 1 } & = 4 \\
u _ { n + 1 } & = \frac { 2 u _ { n } } { 3 } , \quad n \geqslant 1
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item Find the exact values of $u _ { 2 } , u _ { 3 }$ and $u _ { 4 }$
\item Find the value of $u _ { 20 }$, giving your answer to 3 significant figures.
\item Evaluate
$$12 - \sum _ { i = 1 } ^ { 16 } u _ { i }$$
giving your answer to 3 significant figures.
\item Explain why $\sum _ { i = 1 } ^ { N } u _ { i } < 12$ for all positive integer values of $N$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2015 Q10 [8]}}