OCR MEI C3 2011 June — Question 7 4 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2011
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof
TypeDivisibility proof for all integers
DifficultyStandard +0.3 Part (i) is a straightforward difference of two squares requiring simple algebraic manipulation. Part (ii) requires recognizing that 3^(2n) - 1 = (3^n + 1)(3^n - 1) and showing both factors contribute factors of 2, which is a standard divisibility proof technique but requires some insight about odd powers of 3. The total of 4 marks and the 'hence' structure make this slightly easier than average, as the path is clearly signposted.
Spec1.01a Proof: structure of mathematical proof and logical steps

  1. Multiply out \((3^n + 1)(3^n - 1)\). [1]
  2. Hence prove that if \(n\) is a positive integer then \(3^{2n} - 1\) is divisible by 8. [3]

\begin{enumerate}[label=(\roman*)]
\item Multiply out $(3^n + 1)(3^n - 1)$. [1]

\item Hence prove that if $n$ is a positive integer then $3^{2n} - 1$ is divisible by 8. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C3 2011 Q7 [4]}}