Question 6 - A-Level Maths

OCR MEI FP1 2010 January — Question 6 6 marks

Exam Board	OCR MEI
Module	FP1 (Further Pure Mathematics 1)
Year	2010
Session	January
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove summation formula
Difficulty	Standard +0.3 This is a straightforward proof by induction of a summation formula with a standard structure. While it's a Further Maths topic (making it slightly above average), the algebraic manipulation required is routine: verify base case n=1, assume for n=k, then show for n=k+1 by adding (k+1)(k+2) and factoring. No novel insight needed, just following the standard induction template with moderate algebraic facility.
Spec	4.01a Mathematical induction: construct proofs

6 Prove by induction that \(1 \times 2 + 2 \times 3 + \ldots + n ( n + 1 ) = \frac { n ( n + 1 ) ( n + 2 ) } { 3 }\) for all positive integers \(n\). Section B (36 marks)

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Question 6:

Answer	Marks	Guidance
Answer	Mark	Guidance
When \(n=1\), \(\frac{n(n+1)(n+2)}{3} = 2\)	B1
Assume true for \(n=k\)	E1
\(2+6+\cdots+k(k+1) = \frac{k(k+1)(k+2)}{3}\)
\(\Rightarrow 2+6+\cdots+(k+1)(k+2) = \frac{k(k+1)(k+2)}{3}+(k+1)(k+2)\)	M1	Add \((k+1)\)th term to both sides
\(= \frac{1}{3}(k+1)(k+2)(k+3)\)	A1	c.a.o. with correct simplification
\(= \frac{(k+1)((k+1)+1)((k+1)+2)}{3}\)
True for \(n=k \Rightarrow\) true for \(n=k+1\)	E1	Dependent on A1 and previous E1
True for \(n=1\), therefore true for all positive integers	E1 [6]	Dependent on B1 and previous E1

# Question 6:

| Answer | Mark | Guidance |
|--------|------|----------|
| When $n=1$, $\frac{n(n+1)(n+2)}{3} = 2$ | B1 | |
| Assume true for $n=k$ | E1 | |
| $2+6+\cdots+k(k+1) = \frac{k(k+1)(k+2)}{3}$ | | |
| $\Rightarrow 2+6+\cdots+(k+1)(k+2) = \frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$ | M1 | Add $(k+1)$th term to both sides |
| $= \frac{1}{3}(k+1)(k+2)(k+3)$ | A1 | c.a.o. with correct simplification |
| $= \frac{(k+1)((k+1)+1)((k+1)+2)}{3}$ | | |
| True for $n=k \Rightarrow$ true for $n=k+1$ | E1 | Dependent on A1 and previous E1 |
| True for $n=1$, therefore true for all positive integers | E1 **[6]** | Dependent on B1 and previous E1 |

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6 Prove by induction that $1 \times 2 + 2 \times 3 + \ldots + n ( n + 1 ) = \frac { n ( n + 1 ) ( n + 2 ) } { 3 }$ for all positive integers $n$.

Section B (36 marks)\\

\hfill \mbox{\textit{OCR MEI FP1 2010 Q6 [6]}}

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Q1 5 Q2 7 Q3 6 Q4 6 Q5 6 Q6 6 Q7 12 Q8 12 Q9 12