AQA Further AS Paper 1 2018 June — Question 15 4 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2018
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeMethod of differences with given identity
DifficultyStandard +0.3 Part (a) is trivial algebraic verification (1 mark). Part (b) is a standard method of differences application with telescoping seriesβ€”a routine Further Maths technique requiring no novel insight, just careful bookkeeping of terms and simplification of the resulting expression.
Spec4.06b Method of differences: telescoping series
  1. Show that $$\frac{1}{r + 2} - \frac{1}{r + 3} = \frac{1}{(r + 2)(r + 3)}$$ [1 mark]
  2. Use the method of differences to show that $$\sum_{r=1}^{n} \frac{1}{(r + 2)(r + 3)} = \frac{n}{3(n + 3)}$$ [3 marks]
Question 15:
AnswerMarks Guidance
15(a)Shows the result is true with at least one intermediate step. 1.1b
βˆ’ =
π‘Ÿπ‘Ÿ+2 π‘Ÿπ‘Ÿ+3 (π‘Ÿπ‘Ÿ+2)( π‘Ÿπ‘Ÿ+3)
1
=
(π‘Ÿπ‘Ÿ+2)(π‘Ÿπ‘Ÿ+3)
𝑛𝑛 𝑛𝑛
1 1 1
οΏ½οΏ½ οΏ½= οΏ½οΏ½ βˆ’ οΏ½
π‘Ÿπ‘Ÿ=1 (π‘Ÿπ‘Ÿ+2)(π‘Ÿπ‘Ÿ+3) π‘Ÿπ‘Ÿ=1 π‘Ÿπ‘Ÿ+2 π‘Ÿπ‘Ÿ+3
1 1 1 1 1 1
= οΏ½ βˆ’ οΏ½ + οΏ½ βˆ’ οΏ½ + οΏ½ βˆ’ οΏ½ + β‹―β‹―
3 4 4 5 5 6
1 1 1 1
+ οΏ½ βˆ’ οΏ½+οΏ½ βˆ’ οΏ½
𝑛𝑛+1 𝑛𝑛+2 𝑛𝑛+2 𝑛𝑛+3
1 1
= βˆ’
3 𝑛𝑛+3
𝑛𝑛+3βˆ’3
=
3(𝑛𝑛+3)
𝑛𝑛
=
AG 3(𝑛𝑛+3)
AnswerMarks
15(b)Writes at least three corresponding terms of and
1 1
π‘Ÿπ‘Ÿ+2 π‘Ÿπ‘Ÿ+3
Must include the 1st and nth terms and at least the 2nd term or the th term.
AnswerMarks Guidance
(π‘›π‘›βˆ’1)1.1a M1
Correctly uses the method of differences to reduce the sum to two terms.1.1b A1
Completes fully correct proof to reach the required result.
AnswerMarks Guidance
This mark is only available if all previous marks have been awarded.2.1 R1
Total4
QMarking instructions AO 
Question 15:
--- 15(a) ---
15(a) | Shows the result is true with at least one intermediate step. | 1.1b | B1 | 1 1 π‘Ÿπ‘Ÿ+3βˆ’(π‘Ÿπ‘Ÿ+2)
βˆ’ =
π‘Ÿπ‘Ÿ+2 π‘Ÿπ‘Ÿ+3 (π‘Ÿπ‘Ÿ+2)( π‘Ÿπ‘Ÿ+3)
1
=
(π‘Ÿπ‘Ÿ+2)(π‘Ÿπ‘Ÿ+3)
𝑛𝑛 𝑛𝑛
1 1 1
οΏ½οΏ½ οΏ½= οΏ½οΏ½ βˆ’ οΏ½
π‘Ÿπ‘Ÿ=1 (π‘Ÿπ‘Ÿ+2)(π‘Ÿπ‘Ÿ+3) π‘Ÿπ‘Ÿ=1 π‘Ÿπ‘Ÿ+2 π‘Ÿπ‘Ÿ+3
1 1 1 1 1 1
= οΏ½ βˆ’ οΏ½ + οΏ½ βˆ’ οΏ½ + οΏ½ βˆ’ οΏ½ + β‹―β‹―
3 4 4 5 5 6
1 1 1 1
+ οΏ½ βˆ’ οΏ½+οΏ½ βˆ’ οΏ½
𝑛𝑛+1 𝑛𝑛+2 𝑛𝑛+2 𝑛𝑛+3
1 1
= βˆ’
3 𝑛𝑛+3
𝑛𝑛+3βˆ’3
=
3(𝑛𝑛+3)
𝑛𝑛
=
AG 3(𝑛𝑛+3)
--- 15(b) ---
15(b) | Writes at least three corresponding terms of and
1 1
π‘Ÿπ‘Ÿ+2 π‘Ÿπ‘Ÿ+3
Must include the 1st and nth terms and at least the 2nd term or the th term.
(π‘›π‘›βˆ’1) | 1.1a | M1
Correctly uses the method of differences to reduce the sum to two terms. | 1.1b | A1
Completes fully correct proof to reach the required result.
This mark is only available if all previous marks have been awarded. | 2.1 | R1
Total | 4
Q | Marking instructions | AO | Mark | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac{1}{r + 2} - \frac{1}{r + 3} = \frac{1}{(r + 2)(r + 3)}$$
[1 mark]

\item Use the method of differences to show that
$$\sum_{r=1}^{n} \frac{1}{(r + 2)(r + 3)} = \frac{n}{3(n + 3)}$$
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2018 Q15 [4]}}
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