Question 5 - A-Level Maths

AQA FP2 2008 January — Question 5 7 marks

Exam Board	AQA
Module	FP2 (Further Pure Mathematics 2)
Year	2008
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove summation formula
Difficulty	Standard +0.8 This is a Further Maths induction proof involving factorials and a non-standard summation. While the inductive step follows a clear structure, students must carefully manipulate factorial expressions and expand (n+1)² correctly. The algebraic manipulation is more demanding than typical A-level induction proofs, but remains within standard FP2 expectations.
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 $n \geqslant 1$ $$\sum _ { r = 1 } ^ { n } \left( r ^ { 2 } + 1 \right) ( r ! ) = n ( n + 1 ) !$$

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Answer	Marks	Guidance
Assume result true for $n = k$
Then $\sum_{r=1}^{k+1} (r^2 + 1)r!$	M1A1
$= ((k+1)^2 + 1)(k+1)! + k(k+1)!$
Taking out $(k+1)!$ as factor	m1
$= (k+1)!(k^2 + 2k + 1 + 1 + k)$	A1
$= (k+1)!(k+2)!$	A1
$k = 1$ shown $(1^2 + 1)1! = 2$, $1 \times 2! = 2$	B1
$P_k \Rightarrow P_{k+1}$ and $P_1$ true	E1	7 marks; If all 6 marks earned

Question 5 Total: 7 marks

Assume result true for $n = k$ | | 
Then $\sum_{r=1}^{k+1} (r^2 + 1)r!$ | M1A1 | 
$= ((k+1)^2 + 1)(k+1)! + k(k+1)!$ | | 
Taking out $(k+1)!$ as factor | m1 | 
$= (k+1)!(k^2 + 2k + 1 + 1 + k)$ | A1 | 
$= (k+1)!(k+2)!$ | A1 | 
$k = 1$ shown $(1^2 + 1)1! = 2$, $1 \times 2! = 2$ | B1 | 
$P_k \Rightarrow P_{k+1}$ and $P_1$ true | E1 | 7 marks; If all 6 marks earned

**Question 5 Total: 7 marks**

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

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

\hfill \mbox{\textit{AQA FP2 2008 Q5 [7]}}

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Q1 8 Q2 9 Q3 11 Q4 14 Q5 7 Q6 14 Q7 12

Answer	Marks	Guidance
Assume result true for \(n = k\)
Then \(\sum_{r=1}^{k+1} (r^2 + 1)r!\)	M1A1
\(= ((k+1)^2 + 1)(k+1)! + k(k+1)!\)
Taking out \((k+1)!\) as factor	m1
\(= (k+1)!(k^2 + 2k + 1 + 1 + k)\)	A1
\(= (k+1)!(k+2)!\)	A1
\(k = 1\) shown \((1^2 + 1)1! = 2\), \(1 \times 2! = 2\)	B1
\(P_k \Rightarrow P_{k+1}\) and \(P_1\) true	E1	7 marks; If all 6 marks earned