| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find year when threshold exceeded |
| Difficulty | Standard +0.3 This is a straightforward geometric sequences question requiring pattern recognition and basic logarithm manipulation. Part (i) involves simple powers of 2, part (ii)(A) is a standard GP sum proof, and part (ii)(B) applies logarithms to solve an inequality—all routine C2 techniques with clear structure and no novel insight required. Slightly easier than average due to the guided nature and standard methods. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| \(1024\) | 2 marks | M1 for number of buds \(= 2^{10}\) s.o.i. |
| Answer | Marks | Guidance |
|---|---|---|
| \(2047\) | 2 marks | M1 for \(1+2+4+...2^{10}\) or for \(2^{11}-1\) or (their 1024) + 512 + 256 +...+ 1 |
| Answer | Marks | Guidance |
|---|---|---|
| No. of nodes \(= 1 + 2 + ... + 2^{n-1}\) s.o.i. | 1 mark | No. of leaves \(= 7 + 14 + ... + 7 \times 2^{k-1}\) |
| \(\frac{7z(2^n-1)}{2-1}\) | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(7(2^n - 1) > 200\,000\) | M1 | |
| \(2^n > \frac{200\,000}{7} + 1\) or \(\frac{200\,007}{7}\) | M1 | Or \(\log 7 + \log 2^n > \log 200\,007\) |
| \(n \log 2 > \log\left(\frac{200\,007}{7}\right)\) and completion to given ans | M1 | |
| \([n=] 15\) c.a.o. | B1 |
## (i) (A)
$1024$ | 2 marks | M1 for number of buds $= 2^{10}$ s.o.i.
## (i) (B)
$2047$ | 2 marks | M1 for $1+2+4+...2^{10}$ or for $2^{11}-1$ or (their 1024) + 512 + 256 +...+ 1
## (ii) (A)
No. of nodes $= 1 + 2 + ... + 2^{n-1}$ s.o.i. | 1 mark | No. of leaves $= 7 + 14 + ... + 7 \times 2^{k-1}$
$\frac{7z(2^n-1)}{2-1}$ | 1 mark |
## (ii) (B)
$7(2^n - 1) > 200\,000$ | M1 |
$2^n > \frac{200\,000}{7} + 1$ or $\frac{200\,007}{7}$ | M1 | Or $\log 7 + \log 2^n > \log 200\,007$
$n \log 2 > \log\left(\frac{200\,007}{7}\right)$ and completion to given ans | M1 |
$[n=] 15$ c.a.o. | B1 |\includegraphics{figure_12}
A branching plant has stems, nodes, leaves and buds.
• There are 7 leaves at each node.
• From each node, 2 new stems grow.
• At the end of each final stem, there is a bud.
Fig. 12 shows one such plant with 3 stages of nodes. It has 15 stems, 7 nodes, 49 leaves and 8 buds.
\begin{enumerate}[label=(\roman*)]
\item One of these plants has 10 stages of nodes.
\begin{enumerate}[label=(\Alph*)]
\item How many buds does it have? [2]
\item How many stems does it have? [2]
\end{enumerate}
\item \begin{enumerate}[label=(\Alph*)]
\item Show that the number of leaves on one of these plants with $n$ stages of nodes is
$$7(2^n - 1).$$ [2]
\item One of these plants has $n$ stages of nodes and more than 200000 leaves. Show that $n$ satisfies the inequality $n > \frac{\log_{10} 200007 - \log_{10} 7}{\log_{10} 2}$. Hence find the least possible value of $n$. [4]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2010 Q12 [10]}}