OCR MEI Paper 3 2023 June — Question 14 3 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
Year2023
SessionJune
Marks3
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Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeFinding constants from given sum formula
DifficultyStandard +0.8 This is an algebraic verification question requiring careful manipulation of a complex expression involving logarithms and rational functions with multiple terms. While it's a 'show that' rather than a derivation, the algebraic complexity and need to combine fractions with different denominators (involving n and n+1) makes it more demanding than standard A-level work, though not requiring deep conceptual insight.
Spec1.06f Laws of logarithms: addition, subtraction, power rules
14 Show that the expression given in line 33 simplifies to \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \frac { 1 } { \mathrm { r } } \approx \ln \mathrm { n } + \frac { 13 } { 24 } + \frac { 6 \mathrm { n } + 5 } { 12 \mathrm { n } ( \mathrm { n } + 1 ) }\), as given in line 34.
Question 14:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{1}{2n}+\frac{1}{2}+\frac{1}{12}-\frac{1}{24}-\frac{1}{12n}+\frac{1}{12(n+1)}\)M1 AO 1.1a. Or better
\(\left[\int_1^n \frac{1}{x}\,dx\right] = \left[\ln x\right]_1^n = \ln n - \ln 1 = \ln n\)B1 AO 1.1. Integral correctly evaluated. \(\ln 1\) or "\(\ln n - 0\)" seen
\(\ln n + \frac{13}{24} + \frac{6(n+1)-1}{12n(n+1)} = \ln n + \frac{13}{24} + \frac{6n+5}{12n(n+1)}\)A1 [3] AO 2.2a. AG. Convincing completion 
## Question 14:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2n}+\frac{1}{2}+\frac{1}{12}-\frac{1}{24}-\frac{1}{12n}+\frac{1}{12(n+1)}$ | M1 | AO 1.1a. Or better |
| $\left[\int_1^n \frac{1}{x}\,dx\right] = \left[\ln x\right]_1^n = \ln n - \ln 1 = \ln n$ | B1 | AO 1.1. Integral correctly evaluated. $\ln 1$ or "$\ln n - 0$" seen |
| $\ln n + \frac{13}{24} + \frac{6(n+1)-1}{12n(n+1)} = \ln n + \frac{13}{24} + \frac{6n+5}{12n(n+1)}$ | A1 [3] | AO 2.2a. AG. Convincing completion |

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14 Show that the expression given in line 33 simplifies to $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \frac { 1 } { \mathrm { r } } \approx \ln \mathrm { n } + \frac { 13 } { 24 } + \frac { 6 \mathrm { n } + 5 } { 12 \mathrm { n } ( \mathrm { n } + 1 ) }$, as given in line 34.

\hfill \mbox{\textit{OCR MEI Paper 3 2023 Q14 [3]}}
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