OCR MEI Paper 3 Specimen — Question 4 3 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
SessionSpecimen
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypeBasic indefinite integration
DifficultyModerate -0.5 This is a straightforward telescoping series problem requiring only basic logarithm laws (ln(a/b) = ln(a) - ln(b)) and recognition of cancellation pattern. While it requires some algebraic manipulation and the 'show that' format demands clear working, it's a standard technique with no conceptual difficulty beyond A-level core content.
Spec1.04g Sigma notation: for sums of series1.06f Laws of logarithms: addition, subtraction, power rules
4 Show that \(\sum _ { r = 1 } ^ { 4 } \ln \frac { r } { r + 1 } = - \ln 5\).
Question 4:
AnswerMarks Guidance
\(\sum_{r=1}^{4} \ln\frac{r}{r+1} = \ln\frac{1}{2} + \ln\frac{2}{3} + \ln\frac{3}{4} + \ln\frac{4}{5}\)B1 AO 1.1
\(= \ln 1 - \ln 2 + \ln 2 - \ln 3 + \ln 3 - \ln 4 + \ln 4 - \ln 5\)M1 AO 3.1a
\(= 0 - \ln 2 + \ln 2 - \ln 3 + \ln 3 - \ln 4 + \ln 4 - \ln 5 = -\ln 5\)E1 AO 2.2a
[3 marks]
## Question 4:

$\sum_{r=1}^{4} \ln\frac{r}{r+1} = \ln\frac{1}{2} + \ln\frac{2}{3} + \ln\frac{3}{4} + \ln\frac{4}{5}$ | B1 | AO 1.1 | soi

$= \ln 1 - \ln 2 + \ln 2 - \ln 3 + \ln 3 - \ln 4 + \ln 4 - \ln 5$ | M1 | AO 3.1a | Use of $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$ **or** $\ln a + \ln b = \ln ab$; $\ln\left(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\right) = \ln\frac{1}{5} = \ln 1 - \ln 5 = -\ln 5$

$= 0 - \ln 2 + \ln 2 - \ln 3 + \ln 3 - \ln 4 + \ln 4 - \ln 5 = -\ln 5$ | E1 | AO 2.2a | AG

**[3 marks]**
4 Show that $\sum _ { r = 1 } ^ { 4 } \ln \frac { r } { r + 1 } = - \ln 5$.

\hfill \mbox{\textit{OCR MEI Paper 3  Q4 [3]}}
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Q1 2 Q2 4 Q3 4 Q4 3 Q5 5 Q6 5 Q7 2 Q8 8 Q9 7 Q10 10 Q11 10 Q12 1 Q13 3 Q14 3 Q15 5 Q16 3