OCR MEI Paper 3 2021 November — Question 6 4 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
Year2021
SessionNovember
Marks4
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Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeSigma notation: direct numerical evaluation
DifficultyModerate -0.8 This is a straightforward sigma notation problem requiring rationalization of the denominator to create a telescoping series. While it requires recognizing the telescoping pattern, the algebraic manipulation is standard (multiply by conjugate), there are only 3 terms to sum, and the question explicitly tells students what to prove, making it easier than average.
Spec1.02b Surds: manipulation and rationalising denominators1.04g Sigma notation: for sums of series
6 In this question you must show detailed reasoning.
Show that \(\sum _ { r = 1 } ^ { 3 } \frac { 1 } { \sqrt { r + 1 } + \sqrt { r } } = 1\).
Question 6:
AnswerMarks Guidance
\(\frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{3}+\sqrt{2}} + \frac{1}{\sqrt{4}+\sqrt{3}}\)B1 (1.1a) Substituting values
\(\frac{\sqrt{2}-1}{2-1} + \frac{\sqrt{3}-\sqrt{2}}{1} + \frac{\sqrt{4}-\sqrt{3}}{1}\)M1 (3.1a) Attempt to rationalise denominator for one term; Either \(\times\frac{\sqrt{2}-1}{\sqrt{2}-1}\) or \(\frac{\sqrt{2}-1}{2-1}\) at least once for M1
A1 (1.1)All correct
\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3} = 1\)A1 (2.1) Convincing completion (AG)
[4 marks]
## Question 6:

$\frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{3}+\sqrt{2}} + \frac{1}{\sqrt{4}+\sqrt{3}}$ | B1 (1.1a) | Substituting values
$\frac{\sqrt{2}-1}{2-1} + \frac{\sqrt{3}-\sqrt{2}}{1} + \frac{\sqrt{4}-\sqrt{3}}{1}$ | M1 (3.1a) | Attempt to rationalise denominator for one term; Either $\times\frac{\sqrt{2}-1}{\sqrt{2}-1}$ or $\frac{\sqrt{2}-1}{2-1}$ at least once for M1
| A1 (1.1) | All correct
$\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3} = 1$ | A1 (2.1) | Convincing completion (AG)
[4 marks]

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6 In this question you must show detailed reasoning.\\
Show that $\sum _ { r = 1 } ^ { 3 } \frac { 1 } { \sqrt { r + 1 } + \sqrt { r } } = 1$.

\hfill \mbox{\textit{OCR MEI Paper 3 2021 Q6 [4]}}
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Q1 5 Q2 2 Q3 7 Q4 3 Q5 8 Q6 4 Q7 3 Q8 3 Q9 11 Q10 9 Q11 5 Q12 3 Q13 3 Q14 5 Q15 4