AQA Further Paper 2 2019 June — Question 14 12 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypePartial fractions then method of differences
DifficultyChallenging +1.8 This is a challenging Further Maths question requiring method of differences with partial fractions, algebraic manipulation to find a closed form, then sophisticated inequality work involving quadratic manipulation and careful algebraic rearrangement. Part (b) particularly demands insight into how to handle the inequality involving the limit and requires completing the square or quadratic formula application in a non-routine context. The 12-mark allocation and multi-step reasoning place it well above average difficulty.
Spec1.04g Sigma notation: for sums of series4.06b Method of differences: telescoping series
Let $$S_n = \sum_{r=1}^{n} \frac{1}{(r+1)(r+3)}$$ where \(n \geq 1\)
  1. Use the method of differences to show that $$S_n = \frac{5n^2 + an}{12(n+b)(n+c)}$$ where \(a\), \(b\) and \(c\) are integers. [6 marks]
  2. Show that, for any number \(k\) greater than \(\frac{12}{5}\), if the difference between \(\frac{5}{12}\) and \(S_n\) is less than \(\frac{1}{k}\), then $$n > \frac{k-5+\sqrt{k^2+1}}{2}$$ [6 marks]
Question 14:
AnswerMarks Guidance
14(a)Uses partial fractions AO3.1a
βˆ’ =
π‘Ÿπ‘Ÿ+1 π‘Ÿπ‘Ÿ+3 (π‘Ÿπ‘Ÿ+1)(π‘Ÿπ‘Ÿ+3)
𝑛𝑛
1 1
∴ 2𝑆𝑆𝑛𝑛 =οΏ½ βˆ’
π‘Ÿπ‘Ÿ=1π‘Ÿπ‘Ÿ+1 π‘Ÿπ‘Ÿ+3
1 1
= βˆ’
2 4
1 1
+ βˆ’
3 5
1 1
+ βˆ’
4 6
+β‹―
1 1
+ βˆ’
π‘›π‘›βˆ’1 𝑛𝑛+1
1 1
+ βˆ’
𝑛𝑛 𝑛𝑛+2
1 1
+ βˆ’
𝑛𝑛+1 𝑛𝑛+3
1 1 1 1
2𝑆𝑆𝑛𝑛 = + βˆ’ βˆ’
2 3 𝑛𝑛+2 𝑛𝑛+3
5(𝑛𝑛+2)(𝑛𝑛+3)βˆ’6(𝑛𝑛+3+𝑛𝑛+2)
=
6(𝑛𝑛+2)(𝑛𝑛+3)
2
5𝑛𝑛 +13𝑛𝑛
𝑆𝑆𝑛𝑛 =
Correctly expresses the
rational function as
AnswerMarks Guidance
partial fractionsAO1.1b A1
Uses the method of
differences, showing at
least the first three and
last two terms (or vice
versa)
(β€œTerm” here means
one fraction minus
AnswerMarks Guidance
another fraction)AO2.5 M1
Correctly uses the
method of differences
to reduce the
expression to four
AnswerMarks Guidance
terms (oe)AO1.1b A1
Correctly expresses
their three- or four-term
answer with a common
AnswerMarks Guidance
denominatorAO1.1a M1
Completes fully correct
working to reach the
AnswerMarks Guidance
required resultAO2.1 R1
(b)Writes down a correct
inequality.
Condone
5 1
AnswerMarks Guidance
𝑛𝑛AO3.1a B1
5 5𝑛𝑛 +13𝑛𝑛 1
βˆ’ <
12 12(𝑛𝑛+2)(𝑛𝑛+3) π‘˜π‘˜
2
5(𝑛𝑛+2)(𝑛𝑛+3)βˆ’(5𝑛𝑛 +13𝑛𝑛) 1
<
12(𝑛𝑛+2)(𝑛𝑛+3) π‘˜π‘˜
12𝑛𝑛+30 1
<
12(𝑛𝑛+2)(𝑛𝑛+3) π‘˜π‘˜
since both denominators are positive.
π‘˜π‘˜(12𝑛𝑛+30)<1 2(𝑛𝑛+2)(𝑛𝑛+3)
2
The eq1u2a𝑛𝑛ti2o+n (60βˆ’12π‘˜π‘˜)𝑛𝑛+(72βˆ’30π‘˜π‘˜)>0
2𝑛𝑛 +(10βˆ’2π‘˜π‘˜)𝑛𝑛+(12βˆ’5π‘˜π‘˜)>0
has a posi2tive and a negative root
(since 2𝑛𝑛 +(10βˆ’) 2π‘˜π‘˜)𝑛𝑛+(12βˆ’5π‘˜π‘˜)=0
Positiv1e2 roβˆ’o5t π‘˜π‘˜<0
οΏ½οΏ½10βˆ’2π‘˜π‘˜οΏ½ 2
2π‘˜π‘˜βˆ’10+ βˆ’8(12βˆ’5π‘˜π‘˜)
=
4
2
2π‘˜π‘˜βˆ’10+√4π‘˜π‘˜ +4
=
4
2
π‘˜π‘˜βˆ’5+βˆšπ‘˜π‘˜ +1
=
Since the root is positiv2e, if
then 2
12𝑛𝑛 +(60βˆ’12π‘˜π‘˜)𝑛𝑛+(72βˆ’30π‘˜π‘˜)>0
2
π‘˜π‘˜βˆ’5+βˆšπ‘˜π‘˜ +1
𝑛𝑛>
2
12βˆ’π‘†π‘† < π‘˜π‘˜
Simplifies the left-hand
side of their inequality
AnswerMarks Guidance
correctlyAO1.1a M1
Rearranges their
inequality to remove
fraction, explaining that
denominators are
AnswerMarks Guidance
positiveAO2.4 M1
Writes their inequality
or related equation in
simplified quadratic
AnswerMarks Guidance
formAO1.1a M1
Obtains a correct root
or roots of the
quadratic equation in
AnswerMarks Guidance
unsimplified formAO1.1a M1
Completes a rigorous
argument to show the
required result.
This must include a
discussion of the signs
of the roots, or other
convincing reason that
AnswerMarks Guidance
the inequality holds.AO2.1 R1
Total12
QMarking Instructions AO 
Question 14:
--- 14(a) ---
14(a) | Uses partial fractions | AO3.1a | M1 | 1 1 2
βˆ’ =
π‘Ÿπ‘Ÿ+1 π‘Ÿπ‘Ÿ+3 (π‘Ÿπ‘Ÿ+1)(π‘Ÿπ‘Ÿ+3)
𝑛𝑛
1 1
∴ 2𝑆𝑆𝑛𝑛 =οΏ½ βˆ’
π‘Ÿπ‘Ÿ=1π‘Ÿπ‘Ÿ+1 π‘Ÿπ‘Ÿ+3
1 1
= βˆ’
2 4
1 1
+ βˆ’
3 5
1 1
+ βˆ’
4 6
+β‹―
1 1
+ βˆ’
π‘›π‘›βˆ’1 𝑛𝑛+1
1 1
+ βˆ’
𝑛𝑛 𝑛𝑛+2
1 1
+ βˆ’
𝑛𝑛+1 𝑛𝑛+3
1 1 1 1
2𝑆𝑆𝑛𝑛 = + βˆ’ βˆ’
2 3 𝑛𝑛+2 𝑛𝑛+3
5(𝑛𝑛+2)(𝑛𝑛+3)βˆ’6(𝑛𝑛+3+𝑛𝑛+2)
=
6(𝑛𝑛+2)(𝑛𝑛+3)
2
5𝑛𝑛 +13𝑛𝑛
𝑆𝑆𝑛𝑛 =
Correctly expresses the
rational function as
partial fractions | AO1.1b | A1
Uses the method of
differences, showing at
least the first three and
last two terms (or vice
versa)
(β€œTerm” here means
one fraction minus
another fraction) | AO2.5 | M1
Correctly uses the
method of differences
to reduce the
expression to four
terms (oe) | AO1.1b | A1
Correctly expresses
their three- or four-term
answer with a common
denominator | AO1.1a | M1
Completes fully correct
working to reach the
required result | AO2.1 | R1
(b) | Writes down a correct
inequality.
Condone
5 1
𝑛𝑛 | AO3.1a | B1 | 2
5 5𝑛𝑛 +13𝑛𝑛 1
βˆ’ <
12 12(𝑛𝑛+2)(𝑛𝑛+3) π‘˜π‘˜
2
5(𝑛𝑛+2)(𝑛𝑛+3)βˆ’(5𝑛𝑛 +13𝑛𝑛) 1
<
12(𝑛𝑛+2)(𝑛𝑛+3) π‘˜π‘˜
12𝑛𝑛+30 1
<
12(𝑛𝑛+2)(𝑛𝑛+3) π‘˜π‘˜
since both denominators are positive.
π‘˜π‘˜(12𝑛𝑛+30)<1 2(𝑛𝑛+2)(𝑛𝑛+3)
2
The eq1u2a𝑛𝑛ti2o+n (60βˆ’12π‘˜π‘˜)𝑛𝑛+(72βˆ’30π‘˜π‘˜)>0
2𝑛𝑛 +(10βˆ’2π‘˜π‘˜)𝑛𝑛+(12βˆ’5π‘˜π‘˜)>0
has a posi2tive and a negative root
(since 2𝑛𝑛 +(10βˆ’) 2π‘˜π‘˜)𝑛𝑛+(12βˆ’5π‘˜π‘˜)=0
Positiv1e2 roβˆ’o5t π‘˜π‘˜<0
οΏ½οΏ½10βˆ’2π‘˜π‘˜οΏ½ 2
2π‘˜π‘˜βˆ’10+ βˆ’8(12βˆ’5π‘˜π‘˜)
=
4
2
2π‘˜π‘˜βˆ’10+√4π‘˜π‘˜ +4
=
4
2
π‘˜π‘˜βˆ’5+βˆšπ‘˜π‘˜ +1
=
Since the root is positiv2e, if
then 2
12𝑛𝑛 +(60βˆ’12π‘˜π‘˜)𝑛𝑛+(72βˆ’30π‘˜π‘˜)>0
2
π‘˜π‘˜βˆ’5+βˆšπ‘˜π‘˜ +1
𝑛𝑛>
2
12βˆ’π‘†π‘† < π‘˜π‘˜
Simplifies the left-hand
side of their inequality
correctly | AO1.1a | M1
Rearranges their
inequality to remove
fraction, explaining that
denominators are
positive | AO2.4 | M1
Writes their inequality
or related equation in
simplified quadratic
form | AO1.1a | M1
Obtains a correct root
or roots of the
quadratic equation in
unsimplified form | AO1.1a | M1
Completes a rigorous
argument to show the
required result.
This must include a
discussion of the signs
of the roots, or other
convincing reason that
the inequality holds. | AO2.1 | R1
Total | 12
Q | Marking Instructions | AO | Marks | Typical Solution
Let
$$S_n = \sum_{r=1}^{n} \frac{1}{(r+1)(r+3)}$$
where $n \geq 1$

\begin{enumerate}[label=(\alph*)]
\item Use the method of differences to show that
$$S_n = \frac{5n^2 + an}{12(n+b)(n+c)}$$
where $a$, $b$ and $c$ are integers.
[6 marks]

\item Show that, for any number $k$ greater than $\frac{12}{5}$, if the difference between $\frac{5}{12}$ and $S_n$ is less than $\frac{1}{k}$, then
$$n > \frac{k-5+\sqrt{k^2+1}}{2}$$
[6 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2019 Q14 [12]}}
This paper (15 questions)
View full paper
Q1 1 Q2 1 Q3 1 Q4 3 Q5 4 Q6 6 Q7 6 Q8 9 Q9 13 Q10 7 Q11 8 Q12 5 Q13 10 Q14 12 Q15 14