OCR Further Pure Core 2 2021 November — Question 9 6 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2021
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeSuggest and prove formula
DifficultyChallenging +1.2 This is a standard Further Maths induction question involving matrix powers. Students compute the first few powers to spot the pattern (top-right entry is 3n·2^(n-1)), then prove by induction. While it requires matrix multiplication and careful algebraic manipulation, the structure is routine and the pattern recognition is straightforward for Further Maths students. It's moderately above average difficulty due to the algebraic complexity but follows a well-practiced template.
Spec4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar
9 The matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { l l } 2 & 3 \\ 0 & 2 \end{array} \right)\).
  1. By considering \(\mathbf { A } , \mathbf { A } ^ { 2 } , \mathbf { A } ^ { 3 }\) and \(\mathbf { A } ^ { 4 }\) make a conjecture about the form of the matrix \(\mathbf { A } ^ { n }\) in terms of \(n\) for \(n \geqslant 1\).
  2. Use induction to prove the conjecture made in part (a).
Question 9:
AnswerMarks Guidance
9(a) 4 12 8 36
A2 =  , A3 =  ,
0 4  0 8 
16 96
A4 =
 
 0 16
2n 3n×2n−1
Conjecture: An = 
AnswerMarks
 0 2n B1
B1
AnswerMarks
[2]2.2a
2.2bBC
Allow this mark for any
conjecture which works for n =
1, 2, 3 and 4.
AnswerMarks Guidance
9(b) Basis case: n = 1:
21 3×1×20 2 3
A1 = =   =A so true
 0 21  0 2
for n = 1
Assume true for n= k
2k 3k×2k−1
ieAk = 
 0 2k 
2k 3k×2k−12 3
Ak+1 =AkA=
 
 0 2k 0 2
2k+1 3×2k +3k×2k 
= 
 0 2k+1 
2k+1 3(k+1)×2k 
= 
 0 2k+1 
So true for n = k⇒ true for n = k + 1. But true
for n = 1.
AnswerMarks
So true for all positive integer nB1
M1
M1
A1
AnswerMarks
[4]2.1
2.1
2.2a
AnswerMarks
2.4Allow this mark even if the
conjecture is wrong, provided
that it works for n = 1
Must have statement in terms of
some other variable than n.
Conjecture need not be correct.
Uses inductive hypothesis
properly & expands
AG. Manipulating terms
correctly and convincingly to
obtain required form. Some
intermediate working must be
seen and a clear conclusion must
be given for the induction
AnswerMarks
process.A formal proof by induction is
required for full marks.
Question 9:
9 | (a) | 4 12 8 36
A2 =  , A3 =  ,
0 4  0 8 
16 96
A4 =
 
 0 16
2n 3n×2n−1
Conjecture: An = 
 0 2n  | B1
B1
[2] | 2.2a
2.2b | BC
Allow this mark for any
conjecture which works for n =
1, 2, 3 and 4.
9 | (b) | Basis case: n = 1:
21 3×1×20 2 3
A1 = =   =A so true
 0 21  0 2
for n = 1
Assume true for n= k
2k 3k×2k−1
ieAk = 
 0 2k 
2k 3k×2k−12 3
Ak+1 =AkA=
 
 0 2k 0 2
2k+1 3×2k +3k×2k 
= 
 0 2k+1 
2k+1 3(k+1)×2k 
= 
 0 2k+1 
So true for n = k⇒ true for n = k + 1. But true
for n = 1.
So true for all positive integer n | B1
M1
M1
A1
[4] | 2.1
2.1
2.2a
2.4 | Allow this mark even if the
conjecture is wrong, provided
that it works for n = 1
Must have statement in terms of
some other variable than n.
Conjecture need not be correct.
Uses inductive hypothesis
properly & expands
AG. Manipulating terms
correctly and convincingly to
obtain required form. Some
intermediate working must be
seen and a clear conclusion must
be given for the induction
process. | A formal proof by induction is
required for full marks.
9 The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { l l } 2 & 3 \\ 0 & 2 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item By considering $\mathbf { A } , \mathbf { A } ^ { 2 } , \mathbf { A } ^ { 3 }$ and $\mathbf { A } ^ { 4 }$ make a conjecture about the form of the matrix $\mathbf { A } ^ { n }$ in terms of $n$ for $n \geqslant 1$.
\item Use induction to prove the conjecture made in part (a).
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2021 Q9 [6]}}
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