Question 7 - A-Level Maths

AQA Further AS Paper 1 2019 June — Question 7 5 marks

Exam Board	AQA
Module	Further AS Paper 1 (Further AS Paper 1)
Year	2019
Session	June
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Sequences and series, recurrence and convergence
Type	Partial fractions then method of differences
Difficulty	Standard +0.8 This is a standard method of differences question with partial fractions, typical of Further Maths. Part (a) is routine algebraic manipulation (1 mark). Part (b) requires telescoping series technique and algebraic simplification to match the given form, which is methodical but involves careful bookkeeping across multiple steps. While systematic rather than insightful, it's above average difficulty due to the Further Maths context and the algebraic manipulation required to reach the specific form requested.
Spec	4.06b Method of differences: telescoping series

Show that $$\frac{1}{r-1} - \frac{1}{r+1} = \frac{A}{r^2-1}$$ where $A$ is a constant to be found. [1 mark]
Hence use the method of differences to show that $$\sum_{r=2}^n \frac{1}{r^2-1} = \frac{an^2 + bn + c}{4n(n+1)}$$ where $a$, $b$ and $c$ are integers to be found. [4 marks]

Show mark scheme Show mark scheme source

Question 7:

Answer	Marks
7(a)	Obtains the required result with .

Must show at least one intermediate step with no incorrect steps.

𝐴𝐴 = 2

Answer	Marks	Guidance
Condone the LHS not appearing in their working.	1.1b	B1

− ≡ − ≡ 2

Answer	Marks
7(b)	Writes at least five corresponding terms of and

𝑘𝑘 𝑘𝑘

Must include the terms for , , ,

𝑟𝑟−1 𝑟𝑟+1

and for either or

𝑟𝑟 = 2 𝑟𝑟 = 3 𝑟𝑟 =𝑛𝑛−1 𝑟𝑟 = 𝑛𝑛

Condone also included.

𝑟𝑟 = 4 𝑟𝑟 = 𝑛𝑛−2

1 1

Answer	Marks	Guidance
0−2	1.1a	M1

𝑛𝑛 𝑛𝑛

2 1 1

( �)� 2( � = �) �( − ) �

= 1 − 𝑟𝑟1=2 +𝑟𝑟 −11− 1𝑟𝑟=2+𝑟𝑟−1 1− 1𝑟𝑟++1 ..........

1 3 2 4 3 5

( ) ( ) ( )

+ 1 − 1 + 1 − 1 + 1 − 1

n−3 n−1 n−2 n n−1 n+1

1 1 1 1

= + − −

1 2 𝑛𝑛 𝑛𝑛+1

𝑛𝑛

1 1 3 (𝑛𝑛+1)+𝑛𝑛

�� 2 � = � − �

𝑟𝑟=2 𝑟𝑟 −1 2 2 𝑛𝑛(𝑛𝑛+1)

3 2𝑛𝑛+1

= −

4 2𝑛𝑛(𝑛𝑛+1)

2 3(𝑛𝑛 +𝑛𝑛)−2(2𝑛𝑛+1)

=

4𝑛𝑛(𝑛𝑛+1)

2 3𝑛𝑛 −𝑛𝑛−2

=

Correctly uses the method of differences to reduce the expression to

Answer	Marks	Guidance
four terms, or equivalent.	1.1b	A1

Multiplies by (may be seen at any stage).

1

Answer	Marks	Guidance
2	1.1a	M1

Completes fully correct working to reach the required result.

Answer	Marks	Guidance
This mark is only available if all previous marks have been awarded.	2.1	R1
Total	5	4𝑛𝑛(𝑛𝑛+1)
Q	Marking instructions	AO

Question 7:
--- 7(a) ---
7(a) | Obtains the required result with .
Must show at least one intermediate step with no incorrect steps.
𝐴𝐴 = 2
Condone the LHS not appearing in their working. | 1.1b | B1 | 1 1 𝑟𝑟+1 𝑟𝑟−1 2
− ≡ − ≡ 2
--- 7(b) ---
7(b) | Writes at least five corresponding terms of and
𝑘𝑘 𝑘𝑘
Must include the terms for , , ,
𝑟𝑟−1 𝑟𝑟+1
and for either or
𝑟𝑟 = 2 𝑟𝑟 = 3 𝑟𝑟 =𝑛𝑛−1 𝑟𝑟 = 𝑛𝑛
Condone also included.
𝑟𝑟 = 4 𝑟𝑟 = 𝑛𝑛−2
1 1
0−2 | 1.1a | M1 | 𝑟𝑟−1 𝑟𝑟+1 (𝑟𝑟−1)(𝑟𝑟+1) (𝑟𝑟−1)(𝑟𝑟+1) 𝑟𝑟 −1
𝑛𝑛 𝑛𝑛
2 1 1
( �)� 2( � = �) �( − ) �
= 1 − 𝑟𝑟1=2 +𝑟𝑟 −11− 1𝑟𝑟=2+𝑟𝑟−1 1− 1𝑟𝑟++1 ..........
1 3 2 4 3 5
( ) ( ) ( )
+ 1 − 1 + 1 − 1 + 1 − 1
n−3 n−1 n−2 n n−1 n+1
1 1 1 1
= + − −
1 2 𝑛𝑛 𝑛𝑛+1
𝑛𝑛
1 1 3 (𝑛𝑛+1)+𝑛𝑛
�� 2 � = � − �
𝑟𝑟=2 𝑟𝑟 −1 2 2 𝑛𝑛(𝑛𝑛+1)
3 2𝑛𝑛+1
= −
4 2𝑛𝑛(𝑛𝑛+1)
2
3(𝑛𝑛 +𝑛𝑛)−2(2𝑛𝑛+1)
=
4𝑛𝑛(𝑛𝑛+1)
2
3𝑛𝑛 −𝑛𝑛−2
=
Correctly uses the method of differences to reduce the expression to
four terms, or equivalent. | 1.1b | A1
Multiplies by (may be seen at any stage).
1
2 | 1.1a | M1
Completes fully correct working to reach the required result.
This mark is only available if all previous marks have been awarded. | 2.1 | R1
Total | 5 | 4𝑛𝑛(𝑛𝑛+1)
Q | Marking instructions | AO | Marks | Typical solution

Show LaTeX source

\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac{1}{r-1} - \frac{1}{r+1} = \frac{A}{r^2-1}$$
where $A$ is a constant to be found. [1 mark]

\item Hence use the method of differences to show that
$$\sum_{r=2}^n \frac{1}{r^2-1} = \frac{an^2 + bn + c}{4n(n+1)}$$
where $a$, $b$ and $c$ are integers to be found. [4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2019 Q7 [5]}}

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