AQA Further AS Paper 1 2018 June — Question 10 8 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve summation formula
DifficultyStandard +0.8 Part (a) is a standard proof by induction for sum of cubes (routine Further Maths content), but part (b) requires algebraic manipulation to connect r(r-1)(r+1) = rΒ³ - r to the given formula, then apply the result from (a) along with the sum of first n natural numbers. The multi-step algebraic reasoning and the 'hence' connection elevate this above average difficulty, though it remains a recognizable Further Maths exercise type.
Spec4.01a Mathematical induction: construct proofs4.06b Method of differences: telescoping series
  1. Prove by induction that, for all integers \(n \geq 1\), $$\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2$$ [4 marks]
  2. Hence show that $$\sum_{r=1}^{2n} r(r - 1)(r + 1) = n(n + 1)(2n - 1)(2n + 1)$$ [4 marks]
Question 10:
AnswerMarks Guidance
10(a)Demonstrates the rule is correct for
𝑛𝑛 = 11.1b B1
3 3 1 2 2
οΏ½π‘Ÿπ‘Ÿ =1 =1 and Γ—1 Γ—2 =1
π‘Ÿπ‘Ÿ= 1 it is true for 4
∴ 𝑛𝑛 = 1
Assume it is true for
𝑛𝑛 = π‘˜π‘˜
π‘˜π‘˜
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = π‘˜π‘˜ (π‘˜π‘˜+1)
π‘Ÿπ‘Ÿ=1 4
π‘˜π‘˜
3 3 1 2 2 3
οΏ½π‘Ÿπ‘Ÿ +(π‘˜π‘˜+1) = π‘˜π‘˜ (π‘˜π‘˜+1) +(π‘˜π‘˜+1)
π‘Ÿπ‘Ÿ=1 4
π‘˜π‘˜+1
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = (π‘˜π‘˜+1) οΏ½π‘˜π‘˜ +4(π‘˜π‘˜+1)οΏ½
π‘Ÿπ‘Ÿ=1 4
π‘˜π‘˜+1
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = (π‘˜π‘˜+1) (π‘˜π‘˜ +4π‘˜π‘˜+4)
π‘Ÿπ‘Ÿ=1 4
π‘˜π‘˜+1
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = (π‘˜π‘˜+1) (π‘˜π‘˜+2)
π‘Ÿπ‘Ÿ=1it is al 4 so true for
∴T r ue for , and𝑛𝑛 tr=ueπ‘˜π‘˜ fo+r 1 true for , then
by induction it is true for all integers
𝑛𝑛 = 1 𝑛𝑛 = π‘˜π‘˜β‡’ 𝑛𝑛 = π‘˜π‘˜+1
AG 𝑛𝑛 β‰₯ 1
States the rule is true for and adds to
3 1 2 2
AnswerMarks Guidance
𝑛𝑛 = π‘˜π‘˜ (π‘˜π‘˜+1) 4π‘˜π‘˜ (π‘˜π‘˜+1)2.4 M1
Obtains from
1 2 2 1 2 2 3
AnswerMarks Guidance
4(π‘˜π‘˜+1) (π‘˜π‘˜+2) 4π‘˜π‘˜ (π‘˜π‘˜+1) +(π‘˜π‘˜+1)2.2a A1
Completes a rigorous argument and explains how their argument
proves the required result,
AnswerMarks Guidance
This mark is only available if all previous marks have been awarded.2.1 R1
QMarking instructions AO
AnswerMarks
10(b)Expresses LHS as summations of and
Ignore limits of the sums in this part o3nly.
AnswerMarks Guidance
π‘Ÿπ‘Ÿ π‘Ÿπ‘Ÿ1.1b B1
3
οΏ½π‘Ÿπ‘Ÿ(π‘Ÿπ‘Ÿβˆ’1)(π‘Ÿπ‘Ÿ+1)= οΏ½(π‘Ÿπ‘Ÿ βˆ’π‘Ÿπ‘Ÿ)
π‘Ÿπ‘Ÿ=1 π‘Ÿπ‘Ÿ=1
2𝑛𝑛 2𝑛𝑛
3
= οΏ½π‘Ÿπ‘Ÿ βˆ’οΏ½π‘Ÿπ‘Ÿ
π‘Ÿπ‘Ÿ=1 π‘Ÿπ‘Ÿ=1
1 2 2 1
= (2𝑛𝑛) (2𝑛𝑛+1) βˆ’ 2𝑛𝑛(2𝑛𝑛+1)
4 2
2 2
= 𝑛𝑛 (2𝑛𝑛+1) βˆ’π‘›π‘›(2𝑛𝑛+1)
= 𝑛𝑛(2𝑛𝑛+1)(𝑛𝑛(2𝑛𝑛+1)βˆ’1)
2
= 𝑛𝑛(2𝑛𝑛+1)(2𝑛𝑛 +π‘›π‘›βˆ’1)
= 𝑛𝑛(2𝑛𝑛+1)(2π‘›π‘›βˆ’1)(𝑛𝑛+1)
=AG𝑛𝑛 (𝑛𝑛+1)(2π‘›π‘›βˆ’1)(2𝑛𝑛+1)
Expresses LHS in terms of , using part (a) and
AnswerMarks Guidance
11.1a M1
𝑛𝑛 βˆ‘π‘Ÿπ‘Ÿ = 2𝑛𝑛(𝑛𝑛+1)
Takes out as a factor or obtains
Allow one slip in second bracket or one incorr4ect ter3m in 2the expansion.
AnswerMarks Guidance
𝑛𝑛(2𝑛𝑛+1) 4𝑛𝑛 +4𝑛𝑛 βˆ’π‘›π‘› βˆ’π‘›π‘›1.1a M1
Completes fully correct proof to reach the required result.
This mark is only available if all previous marks have been awarded.
Note:
4 3 2
AnswerMarks Guidance
𝑛𝑛(𝑛𝑛+1)(2π‘›π‘›βˆ’1)(2𝑛𝑛+1)= 4𝑛𝑛 +4𝑛𝑛 βˆ’π‘›π‘› βˆ’π‘›π‘›2.1 R1
Total8
QMarking instructions AO 
Question 10:
--- 10(a) ---
10(a) | Demonstrates the rule is correct for
𝑛𝑛 = 1 | 1.1b | B1 | 1
3 3 1 2 2
οΏ½π‘Ÿπ‘Ÿ =1 =1 and Γ—1 Γ—2 =1
π‘Ÿπ‘Ÿ= 1 it is true for 4
∴ 𝑛𝑛 = 1
Assume it is true for
𝑛𝑛 = π‘˜π‘˜
π‘˜π‘˜
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = π‘˜π‘˜ (π‘˜π‘˜+1)
π‘Ÿπ‘Ÿ=1 4
π‘˜π‘˜
3 3 1 2 2 3
οΏ½π‘Ÿπ‘Ÿ +(π‘˜π‘˜+1) = π‘˜π‘˜ (π‘˜π‘˜+1) +(π‘˜π‘˜+1)
π‘Ÿπ‘Ÿ=1 4
π‘˜π‘˜+1
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = (π‘˜π‘˜+1) οΏ½π‘˜π‘˜ +4(π‘˜π‘˜+1)οΏ½
π‘Ÿπ‘Ÿ=1 4
π‘˜π‘˜+1
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = (π‘˜π‘˜+1) (π‘˜π‘˜ +4π‘˜π‘˜+4)
π‘Ÿπ‘Ÿ=1 4
π‘˜π‘˜+1
3 1 2 2
οΏ½π‘Ÿπ‘Ÿ = (π‘˜π‘˜+1) (π‘˜π‘˜+2)
π‘Ÿπ‘Ÿ=1it is al 4 so true for
∴T r ue for , and𝑛𝑛 tr=ueπ‘˜π‘˜ fo+r 1 true for , then
by induction it is true for all integers
𝑛𝑛 = 1 𝑛𝑛 = π‘˜π‘˜β‡’ 𝑛𝑛 = π‘˜π‘˜+1
AG 𝑛𝑛 β‰₯ 1
States the rule is true for and adds to
3 1 2 2
𝑛𝑛 = π‘˜π‘˜ (π‘˜π‘˜+1) 4π‘˜π‘˜ (π‘˜π‘˜+1) | 2.4 | M1
Obtains from
1 2 2 1 2 2 3
4(π‘˜π‘˜+1) (π‘˜π‘˜+2) 4π‘˜π‘˜ (π‘˜π‘˜+1) +(π‘˜π‘˜+1) | 2.2a | A1
Completes a rigorous argument and explains how their argument
proves the required result,
This mark is only available if all previous marks have been awarded. | 2.1 | R1
Q | Marking instructions | AO | Mark | Typical solution
--- 10(b) ---
10(b) | Expresses LHS as summations of and
Ignore limits of the sums in this part o3nly.
π‘Ÿπ‘Ÿ π‘Ÿπ‘Ÿ | 1.1b | B1 | 2𝑛𝑛 2𝑛𝑛
3
οΏ½π‘Ÿπ‘Ÿ(π‘Ÿπ‘Ÿβˆ’1)(π‘Ÿπ‘Ÿ+1)= οΏ½(π‘Ÿπ‘Ÿ βˆ’π‘Ÿπ‘Ÿ)
π‘Ÿπ‘Ÿ=1 π‘Ÿπ‘Ÿ=1
2𝑛𝑛 2𝑛𝑛
3
= οΏ½π‘Ÿπ‘Ÿ βˆ’οΏ½π‘Ÿπ‘Ÿ
π‘Ÿπ‘Ÿ=1 π‘Ÿπ‘Ÿ=1
1 2 2 1
= (2𝑛𝑛) (2𝑛𝑛+1) βˆ’ 2𝑛𝑛(2𝑛𝑛+1)
4 2
2 2
= 𝑛𝑛 (2𝑛𝑛+1) βˆ’π‘›π‘›(2𝑛𝑛+1)
= 𝑛𝑛(2𝑛𝑛+1)(𝑛𝑛(2𝑛𝑛+1)βˆ’1)
2
= 𝑛𝑛(2𝑛𝑛+1)(2𝑛𝑛 +π‘›π‘›βˆ’1)
= 𝑛𝑛(2𝑛𝑛+1)(2π‘›π‘›βˆ’1)(𝑛𝑛+1)
=AG𝑛𝑛 (𝑛𝑛+1)(2π‘›π‘›βˆ’1)(2𝑛𝑛+1)
Expresses LHS in terms of , using part (a) and
1 | 1.1a | M1
𝑛𝑛 βˆ‘π‘Ÿπ‘Ÿ = 2𝑛𝑛(𝑛𝑛+1)
Takes out as a factor or obtains
Allow one slip in second bracket or one incorr4ect ter3m in 2the expansion.
𝑛𝑛(2𝑛𝑛+1) 4𝑛𝑛 +4𝑛𝑛 βˆ’π‘›π‘› βˆ’π‘›π‘› | 1.1a | M1
Completes fully correct proof to reach the required result.
This mark is only available if all previous marks have been awarded.
Note:
4 3 2
𝑛𝑛(𝑛𝑛+1)(2π‘›π‘›βˆ’1)(2𝑛𝑛+1)= 4𝑛𝑛 +4𝑛𝑛 βˆ’π‘›π‘› βˆ’π‘›π‘› | 2.1 | R1
Total | 8
Q | Marking instructions | AO | Mark | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Prove by induction that, for all integers $n \geq 1$,
$$\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2$$
[4 marks]

\item Hence show that
$$\sum_{r=1}^{2n} r(r - 1)(r + 1) = n(n + 1)(2n - 1)(2n + 1)$$
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2018 Q10 [8]}}
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