AQA Further AS Paper 1 2020 June — Question 5 4 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2020
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeProving standard summation formulae
DifficultyStandard +0.3 Part (a) is straightforward algebra requiring factorization of rΒ²[(r+1)Β² - (r-1)Β²] to find p=4. Part (b) applies the standard method of differences technique with telescoping sums, which is a routine Further Maths topic. While it requires understanding of summation notation and the telescoping mechanism, this is a textbook application with clear scaffolding from part (a), making it easier than average even for Further Maths.
Spec4.06b Method of differences: telescoping series
  1. Show that $$r^2(r + 1)^2 - (r - 1)^2r^2 = pr^3$$ where \(p\) is an integer to be found. [1 mark]
  2. Hence use the method of differences to show that $$\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2$$ [3 marks]
Question 5:
AnswerMarks
5(a)Completes a rigorous argument to show that
2 2 2 2 3
AnswerMarks Guidance
Must show at leaπ‘Ÿπ‘Ÿst (oπ‘Ÿπ‘Ÿn+e i1n)terβˆ’me(dπ‘Ÿπ‘Ÿiaβˆ’te1 s)teπ‘Ÿπ‘Ÿp. = 4π‘Ÿπ‘Ÿ2.1 R1
π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ+1) βˆ’(π‘Ÿπ‘Ÿβˆ’1) π‘Ÿπ‘Ÿ = π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ +2π‘Ÿπ‘Ÿ+1)βˆ’π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ βˆ’2π‘Ÿπ‘Ÿ+1)
4 3 2 4 3 2 3
= π‘Ÿπ‘Ÿ +2π‘Ÿπ‘Ÿ +π‘Ÿπ‘Ÿ βˆ’π‘Ÿπ‘Ÿ +2π‘Ÿπ‘Ÿ βˆ’π‘Ÿπ‘Ÿ =4π‘Ÿπ‘Ÿ
AnswerMarks
5(b)Uses the result from (a) with their to express in terms of
with one pair of terms of3 the sum
𝑝𝑝 βˆ‘π‘Ÿπ‘Ÿ
writ2ten corre2ctly. 2 2
AnswerMarks Guidance
βˆ‘π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ+1) βˆ’(π‘Ÿπ‘Ÿβˆ’1) π‘Ÿπ‘Ÿ1.1a M1
3 2 2 2 2
οΏ½4π‘Ÿπ‘Ÿ = οΏ½[π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ+ 1) βˆ’(π‘Ÿπ‘Ÿβˆ’1) π‘Ÿπ‘Ÿ ]
π‘Ÿπ‘Ÿ=1 π‘Ÿπ‘Ÿ=1
2 2 2 2
= 1 Γ—2 βˆ’0 Γ—1
2 2 2 2
+ 2 Γ—3 βˆ’1 Γ—2
+ ………………
2 2 2 2
+ (π‘›π‘›βˆ’1) 𝑛𝑛 βˆ’(π‘›π‘›βˆ’2) (π‘›π‘›βˆ’1)
2 2 2 2
+ 𝑛𝑛 (𝑛𝑛+1) βˆ’(π‘›π‘›βˆ’1) 𝑛𝑛
2 2 2 2
= 𝑛𝑛 (𝑛𝑛+1) βˆ’0 Γ—1
𝑛𝑛
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = 𝑛𝑛 (𝑛𝑛+1)
Writes down at least three pairs of terms including the first and
AnswerMarks Guidance
last pair of terms of the sum.1.1b A1
Completes a reasoned argument using the method of
differences to show that
𝑛𝑛
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = 𝑛𝑛 (𝑛𝑛+1)
AnswerMarks Guidance
π‘Ÿπ‘Ÿ=1 42.1 R1
Total4 π‘Ÿπ‘Ÿ=1 4
QMarking instructions AO 
Question 5:
--- 5(a) ---
5(a) | Completes a rigorous argument to show that
2 2 2 2 3
Must show at leaπ‘Ÿπ‘Ÿst (oπ‘Ÿπ‘Ÿn+e i1n)terβˆ’me(dπ‘Ÿπ‘Ÿiaβˆ’te1 s)teπ‘Ÿπ‘Ÿp. = 4π‘Ÿπ‘Ÿ | 2.1 | R1 | 2 2 2 2 2 2 2 2
π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ+1) βˆ’(π‘Ÿπ‘Ÿβˆ’1) π‘Ÿπ‘Ÿ = π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ +2π‘Ÿπ‘Ÿ+1)βˆ’π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ βˆ’2π‘Ÿπ‘Ÿ+1)
4 3 2 4 3 2 3
= π‘Ÿπ‘Ÿ +2π‘Ÿπ‘Ÿ +π‘Ÿπ‘Ÿ βˆ’π‘Ÿπ‘Ÿ +2π‘Ÿπ‘Ÿ βˆ’π‘Ÿπ‘Ÿ =4π‘Ÿπ‘Ÿ
--- 5(b) ---
5(b) | Uses the result from (a) with their to express in terms of
with one pair of terms of3 the sum
𝑝𝑝 βˆ‘π‘Ÿπ‘Ÿ
writ2ten corre2ctly. 2 2
βˆ‘π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ+1) βˆ’(π‘Ÿπ‘Ÿβˆ’1) π‘Ÿπ‘Ÿ | 1.1a | M1 | 𝑛𝑛 𝑛𝑛
3 2 2 2 2
οΏ½4π‘Ÿπ‘Ÿ = οΏ½[π‘Ÿπ‘Ÿ (π‘Ÿπ‘Ÿ+ 1) βˆ’(π‘Ÿπ‘Ÿβˆ’1) π‘Ÿπ‘Ÿ ]
π‘Ÿπ‘Ÿ=1 π‘Ÿπ‘Ÿ=1
2 2 2 2
= 1 Γ—2 βˆ’0 Γ—1
2 2 2 2
+ 2 Γ—3 βˆ’1 Γ—2
+ ………………
2 2 2 2
+ (π‘›π‘›βˆ’1) 𝑛𝑛 βˆ’(π‘›π‘›βˆ’2) (π‘›π‘›βˆ’1)
2 2 2 2
+ 𝑛𝑛 (𝑛𝑛+1) βˆ’(π‘›π‘›βˆ’1) 𝑛𝑛
2 2 2 2
= 𝑛𝑛 (𝑛𝑛+1) βˆ’0 Γ—1
𝑛𝑛
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = 𝑛𝑛 (𝑛𝑛+1)
Writes down at least three pairs of terms including the first and
last pair of terms of the sum. | 1.1b | A1
Completes a reasoned argument using the method of
differences to show that
𝑛𝑛
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = 𝑛𝑛 (𝑛𝑛+1)
π‘Ÿπ‘Ÿ=1 4 | 2.1 | R1
Total | 4 | π‘Ÿπ‘Ÿ=1 4
Q | Marking instructions | AO | Marks | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Show that
$$r^2(r + 1)^2 - (r - 1)^2r^2 = pr^3$$
where $p$ is an integer to be found.
[1 mark]

\item Hence use the method of differences to show that
$$\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2$$
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2020 Q5 [4]}}
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