Pre-U Pre-U 9795/1 2015 June — Question 3 6 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2015
SessionJune
Marks6
TopicProof by induction
TypeProve matrix power formula
DifficultyChallenging +1.2 This is a structured proof by induction with matrices, requiring verification of the base case and inductive step through matrix-vector multiplication. While it involves multiple algebraic steps and careful bookkeeping with the quadratic terms, the approach is entirely standard and mechanical once the induction framework is set up. The matrix multiplication itself is routine, making this moderately above average difficulty but not requiring significant insight.
Spec4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar
\(\mathbf{M}\) is the matrix \(\begin{pmatrix} 1 & -2 & 2 \\ 2 & -1 & 2 \\ 2 & -2 & 3 \end{pmatrix}\). Use induction to prove that, for all positive integers \(n\), $$\mathbf{M}^n \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2n + 1 \\ 2n^2 + 2n \\ 2n^2 + 2n + 1 \end{pmatrix}.$$ [6]
Question 3:
AnswerMarks
31 3  2(1)+1 
     
n = 1, M0=4 = 2(1)2 +2(1)  ⇒ result true for n = 1 (both sides shown)
     
1 5 2(1)2 +2(1)+1
     
1  2k+1 
   
Mk0= 2k2 +2k 
induction hypothesis that
   
1 2k2 +2k+1
   
1  2k+1  2k+1−4k2 −4k+4k2 +4k+2
     
attempt at Mk + 10=M 2k2 +2k =4k+2−2k2 −2k+4k2 +4k+2
     
1 2k2 +2k+1 4k+2−4k2 −4k+6k2 +6k+3
     
 2k+3 
 
2k2 +6k+4
=
 
2k2 +6k+5
 
 2(k+1)+1 
 
 2(k+1)2 +2(k+1) 
=
 
2(k+1)2 +2(k+1)+1
 
This is the result with k replaced by (k + 1).
Hence IF the result is true for n = k THEN it also true for n = k + 1.
AnswerMarks
Since the result is true for n = 1, it follows that it is true for all positive integers n.B1
M1
M1
A1
A1
E1
[6]
AnswerMarks Guidance
Page 3Mark Scheme Syllabus
Cambridge Pre-U – May/June 20159795 01 
Question 3:
3 | 1 3  2(1)+1 
     
n = 1, M0=4 = 2(1)2 +2(1)  ⇒ result true for n = 1 (both sides shown)
     
1 5 2(1)2 +2(1)+1
     
1  2k+1 
   
Mk0= 2k2 +2k 
induction hypothesis that
   
1 2k2 +2k+1
   
1  2k+1  2k+1−4k2 −4k+4k2 +4k+2
     
attempt at Mk + 10=M 2k2 +2k =4k+2−2k2 −2k+4k2 +4k+2
     
1 2k2 +2k+1 4k+2−4k2 −4k+6k2 +6k+3
     
 2k+3 
 
2k2 +6k+4
=
 
2k2 +6k+5
 
 2(k+1)+1 
 
 2(k+1)2 +2(k+1) 
=
 
2(k+1)2 +2(k+1)+1
 
This is the result with k replaced by (k + 1).
Hence IF the result is true for n = k THEN it also true for n = k + 1.
Since the result is true for n = 1, it follows that it is true for all positive integers n. | B1
M1
M1
A1
A1
E1
[6]
Page 3 | Mark Scheme | Syllabus | Paper
Cambridge Pre-U – May/June 2015 | 9795 | 01
$\mathbf{M}$ is the matrix $\begin{pmatrix} 1 & -2 & 2 \\ 2 & -1 & 2 \\ 2 & -2 & 3 \end{pmatrix}$. Use induction to prove that, for all positive integers $n$,
$$\mathbf{M}^n \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2n + 1 \\ 2n^2 + 2n \\ 2n^2 + 2n + 1 \end{pmatrix}.$$ [6]

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2015 Q3 [6]}}
This paper (13 questions)
View full paper
Q1 3 Q2 3 Q3 6 Q4 7 Q5 11 Q6 9 Q7 7 Q8 9 Q9 9 Q10 11 Q11 13 Q12 22 Q13 10