OCR MEI Further Extra Pure 2021 November — Question 6 8 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeUse series to approximate numerical value
DifficultyChallenging +1.8 This is a Further Maths proof question requiring manipulation of infinite series, geometric series comparison, and a classic irrationality proof. Part (a) needs insight to bound the series geometrically, part (b) requires understanding divisibility arguments, and part (c) is the standard proof that e is irrational—sophisticated but well-scaffolded with 8 marks total.
Spec4.01b Complex proofs: conjecture and demanding proofs8.01c Sequence behaviour: periodic, convergent, divergent, oscillating, monotonic
You are given that \(q \in \mathbb{Z}\) with \(q \geqslant 1\) and that $$S = \frac{1}{(q+1)} + \frac{1}{(q+1)(q+2)} + \frac{1}{(q+1)(q+2)(q+3)} + \cdots$$
  1. By considering a suitable geometric series show that \(S < \frac{1}{q}\). [3]
  2. Deduce that \(S \notin \mathbb{Z}\). [2]
You are also given that \(\mathrm{e} = \sum_{r=0}^{\infty} \frac{1}{r!}\).
  1. Assume that \(\mathrm{e} = \frac{p}{q}\), where \(p\) and \(q\) are positive integers. By writing the infinite series for \(\mathrm{e}\) in a form using \(q\) and \(S\) and using the result from part (b), prove by contradiction that \(\mathrm{e}\) is irrational. [3]
Question 6:
AnswerMarks Guidance
6(a) 1 1 1
+ + +
(q+1) (q+1)(q+2) (q+1)(q+2)(q+3)
1 1 1
< + + +
(q+1) (q+1)2 (q+1)3
1
q+1
=
1
1−
q+1
1 1
= =
AnswerMarks
q+1−1 qM1
A1
A1
AnswerMarks
[3]2.1
2.1
AnswerMarks
2.1Correct statement that given series
is less than an infinite GP (could
1 1 1 1
be eg + +... or + +...).
q q2 3 32
a
FT on their .
1−r
AG. Intermediate step must be
seen.
AnswerMarks Guidance
6(b) 1
q≥1⇒ ≤1
q
1
But S < ⇒S <1; clearly S > 0 so 0<S <1so
q
AnswerMarks
S∉ .M1
A1
AnswerMarks
[2]2.2a
2.2aAG. S > 0 must be stated but need
not be justified.1 1
Since 0< ≤1 and S < then
q q
0<S <1 and ∴S∉ .
AnswerMarks Guidance
6(c) ∞ ∞
1 p q!
e=∑ = ⇒eq!=∑ = p(q−1)!
r! q r!
r=0 r=0
∞ q ∞
q! q! q!
∴p(q−1)!=∑ =∑ + ∑
r! r! r!
r=0 r=0 r=q+1
q! q!
=q!+q!+ +...+ +S
2! q!
=2q!+q(q−1)...×3+q(q−1)...×4+...+1+S
q!
p(q – 1)! and q!+q!+ +...+1 are all integers
2!
AnswerMarks
but S is not which is a contradiction.M1
M1
AnswerMarks
A13.1a
2.1
AnswerMarks
3.2aMultiplying both sides by q! No
need to mention q ≥ 1 in this part.
Rewriting to a form in which it is
clear that every term on both sides,
except S, is an integer.
AG
AnswerMarks Guidance
Alternative Method:M1 Expressing S as an infinite sum in
terms of factorials.
∞
q!
S = ∑
r!
r=q+1
∞ 1 q q! q q!
S =q!∑ −∑ =q!e−q!−∑
r! r! r!
r=0 r=0 r=1
q
= p(q−1)!−q!−∑q(q−1)...(q−r+1)
r=1
AnswerMarks Guidance
since 1≤r ≤qM1 Rewriting to a form in which it is
clear that every term on both sides,
except S, is an integer.
AnswerMarks Guidance
p(q – 1)!, q! and q(q – 1)...(q – r + 1) are allA1 AG
integers but S is not which is a contradiction.
[3]
M1
Expressing S as an infinite sum in
terms of factorials.
Rewriting to a form in which it is
clear that every term on both sides,
except S, is an integer.
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
Question 6:
6 | (a) | 1 1 1
+ + +
(q+1) (q+1)(q+2) (q+1)(q+2)(q+3)
1 1 1
< + + +
(q+1) (q+1)2 (q+1)3
1
q+1
=
1
1−
q+1
1 1
= =
q+1−1 q | M1
A1
A1
[3] | 2.1
2.1
2.1 | Correct statement that given series
is less than an infinite GP (could
1 1 1 1
be eg + +... or + +...).
q q2 3 32
a
FT on their .
1−r
AG. Intermediate step must be
seen.
6 | (b) | 1
q≥1⇒ ≤1
q
1
But S < ⇒S <1; clearly S > 0 so 0<S <1so
q
S∉ . | M1
A1
[2] | 2.2a
2.2a | AG. S > 0 must be stated but need
not be justified. | 1 1
Since 0< ≤1 and S < then
q q
0<S <1 and ∴S∉ .
6 | (c) | ∞ ∞
1 p q!
e=∑ = ⇒eq!=∑ = p(q−1)!
r! q r!
r=0 r=0
∞ q ∞
q! q! q!
∴p(q−1)!=∑ =∑ + ∑
r! r! r!
r=0 r=0 r=q+1
q! q!
=q!+q!+ +...+ +S
2! q!
=2q!+q(q−1)...×3+q(q−1)...×4+...+1+S
q!
p(q – 1)! and q!+q!+ +...+1 are all integers
2!
but S is not which is a contradiction. | M1
M1
A1 | 3.1a
2.1
3.2a | Multiplying both sides by q! No
need to mention q ≥ 1 in this part.
Rewriting to a form in which it is
clear that every term on both sides,
except S, is an integer.
AG
Alternative Method: | M1 | Expressing S as an infinite sum in
terms of factorials.
∞
q!
S = ∑
r!
r=q+1
∞ 1 q q! q q!
S =q!∑ −∑ =q!e−q!−∑
r! r! r!
r=0 r=0 r=1
q
= p(q−1)!−q!−∑q(q−1)...(q−r+1)
r=1
since 1≤r ≤q | M1 | Rewriting to a form in which it is
clear that every term on both sides,
except S, is an integer.
p(q – 1)!, q! and q(q – 1)...(q – r + 1) are all | A1 | AG
integers but S is not which is a contradiction.
[3]
M1
Expressing S as an infinite sum in
terms of factorials.
Rewriting to a form in which it is
clear that every term on both sides,
except S, is an integer.
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
You are given that $q \in \mathbb{Z}$ with $q \geqslant 1$ and that
$$S = \frac{1}{(q+1)} + \frac{1}{(q+1)(q+2)} + \frac{1}{(q+1)(q+2)(q+3)} + \cdots$$

\begin{enumerate}[label=(\alph*)]
\item By considering a suitable geometric series show that $S < \frac{1}{q}$. [3]

\item Deduce that $S \notin \mathbb{Z}$. [2]
\end{enumerate}

You are also given that $\mathrm{e} = \sum_{r=0}^{\infty} \frac{1}{r!}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Assume that $\mathrm{e} = \frac{p}{q}$, where $p$ and $q$ are positive integers. By writing the infinite series for $\mathrm{e}$ in a form using $q$ and $S$ and using the result from part (b), prove by contradiction that $\mathrm{e}$ is irrational. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure 2021 Q6 [8]}}
This paper (6 questions)
View full paper
Q1 11 Q2 7 Q3 14 Q4 14 Q5 6 Q6 8