Question 5 - A-Level Maths

AQA Further Paper 2 2022 June — Question 5 4 marks

Exam Board	AQA
Module	Further Paper 2 (Further Paper 2)
Year	2022
Session	June
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove summation formula
Difficulty	Standard +0.3 This is a standard proof by induction for a summation formula with a straightforward algebraic structure. The base case is trivial, and the inductive step requires only routine algebraic manipulation of the well-known sum of cubes formula. While it's a Further Maths question, it's one of the most common induction exercises with no novel insight required, making it slightly easier than average overall.
Spec	4.01a Mathematical induction: construct proofs 4.06a Summation formulae: sum of r, r^2, r^3

5 Prove by induction that, for all integers $n \geq 1$, $$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \left\{ \frac { 1 } { 2 } n ( n + 1 ) \right\} ^ { 2 }$$ [4 marks]

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Question 5:

Answer	Marks	Guidance
Base case: Let $n=1$, then $\left(\frac{1}{2}n(n+1)\right)^2 = 1^2 = 1$ and $\sum_{r=1}^{n} r^3 = 1$, so the result is true for $n=1$	B1	Must demonstrate result AND state it is true for $n=1$
Inductive step: Assume the result is true for $n=k$, then $\sum_{r=1}^{k} r^3 = \left(\frac{1}{2}k(k+1)\right)^2$	M1	Must assume true for $n=k$ AND add $(k+1)^3$
\[\sum_{r=1}^{k+1} r^3 = \left(\frac{1}{2}k(k+1)\right)^2 + (k+1)^3 = \frac{1}{4}(k+1)^2(k^2+4(k+1))\] \[= \frac{1}{4}(k+1)^2(k^2+4k+4) = \frac{1}{4}(k+1)^2(k+2)^2 = \left(\frac{1}{2}(k+1)(k+2)\right)^2\]	A1	Must obtain $\frac{1}{4}(k+1)^2(k+2)^2$
Conclusion: The result is true for $n=1$; if true for $n=k$, then it's also true for $n=k+1$ and hence by induction $\sum_{r=1}^{n} r^3 = \left\{\frac{1}{2}n(n+1)\right\}^2$ for all integers $n \geq 1$	R1	Must state both base case and inductive step in conclusion

## Question 5:

**Base case:** Let $n=1$, then $\left(\frac{1}{2}n(n+1)\right)^2 = 1^2 = 1$ and $\sum_{r=1}^{n} r^3 = 1$, so the result is true for $n=1$ | B1 | Must demonstrate result AND state it is true for $n=1$

**Inductive step:** Assume the result is true for $n=k$, then $\sum_{r=1}^{k} r^3 = \left(\frac{1}{2}k(k+1)\right)^2$ | M1 | Must assume true for $n=k$ AND add $(k+1)^3$

$$\sum_{r=1}^{k+1} r^3 = \left(\frac{1}{2}k(k+1)\right)^2 + (k+1)^3 = \frac{1}{4}(k+1)^2(k^2+4(k+1))$$ $$= \frac{1}{4}(k+1)^2(k^2+4k+4) = \frac{1}{4}(k+1)^2(k+2)^2 = \left(\frac{1}{2}(k+1)(k+2)\right)^2$$ | A1 | Must obtain $\frac{1}{4}(k+1)^2(k+2)^2$

**Conclusion:** The result is true for $n=1$; if true for $n=k$, then it's also true for $n=k+1$ and hence by induction $\sum_{r=1}^{n} r^3 = \left\{\frac{1}{2}n(n+1)\right\}^2$ for all integers $n \geq 1$ | R1 | Must state both base case and inductive step in conclusion

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5 Prove by induction that, for all integers $n \geq 1$,

$$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \left\{ \frac { 1 } { 2 } n ( n + 1 ) \right\} ^ { 2 }$$

[4 marks]\\

\hfill \mbox{\textit{AQA Further Paper 2 2022 Q5 [4]}}

This paper (14 questions)

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Q1 1 Q2 1 Q3 1 Q4 1 Q5 4 Q6 3 Q7 8 Q8 10 Q9 14 Q10 7 Q11 9 Q12 11 Q13 16 Q14 14

Answer	Marks	Guidance
Base case: Let \(n=1\), then \(\left(\frac{1}{2}n(n+1)\right)^2 = 1^2 = 1\) and \(\sum_{r=1}^{n} r^3 = 1\), so the result is true for \(n=1\)	B1	Must demonstrate result AND state it is true for \(n=1\)
Inductive step: Assume the result is true for \(n=k\), then \(\sum_{r=1}^{k} r^3 = \left(\frac{1}{2}k(k+1)\right)^2\)	M1	Must assume true for \(n=k\) AND add \((k+1)^3\)
\[\sum_{r=1}^{k+1} r^3 = \left(\frac{1}{2}k(k+1)\right)^2 + (k+1)^3 = \frac{1}{4}(k+1)^2(k^2+4(k+1))\] \[= \frac{1}{4}(k+1)^2(k^2+4k+4) = \frac{1}{4}(k+1)^2(k+2)^2 = \left(\frac{1}{2}(k+1)(k+2)\right)^2\]	A1	Must obtain \(\frac{1}{4}(k+1)^2(k+2)^2\)
Conclusion: The result is true for \(n=1\); if true for \(n=k\), then it's also true for \(n=k+1\) and hence by induction \(\sum_{r=1}^{n} r^3 = \left\{\frac{1}{2}n(n+1)\right\}^2\) for all integers \(n \geq 1\)	R1	Must state both base case and inductive step in conclusion