EXM.2.AHL.TZ0.7

pestleMathematicsAIHLPaper 2EXM· ahl-1-14-introduction-to-matricessource ↗

Let $S$ _$n$ be the sum of the first $n$ terms of the arithmetic series 2 + 4 + 6 + ….

Let M = $\left( {\begin{array}{*{20}{c}} 1&2 \\ 0&1 \end{array}} \right)$ .

It may now be assumed that M^$n$ = $\left( {\begin{array}{*{20}{c}} 1&{2n} \\ 0&1 \end{array}} \right)$ , for $n$ ≥ 4. The sum T_$n$ is defined by

T_$n$ = M¹ + M² + M³ + ... + M^$n$.

Find $S$ ₄.

[1]

a.i.

Find $S$ ₁₀₀.

[3]

a.ii.

Find M².

[2]

b.i.

Show that M³ = $\left( {\begin{array}{*{20}{c}} 1&6 \\ 0&1 \end{array}} \right)$ .

[3]

b.ii.

Write down M⁴.

[1]

c.i.

Find T₄.

[3]

c.ii.

Using your results from part (a) (ii), find T₁₀₀.

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$S$ ₄ = 20 A1 N1

[1 mark]

a.i.

$u$ ₁ = 2, $d$ = 2 (A1)

Attempting to use formula for $S$ _$n$ M1

$S$ ₁₀₀ = 10100 A1 N2

[3 marks]

a.ii.

M² = $\left( {\begin{array}{*{20}{c}} 1&4 \\ 0&1 \end{array}} \right)$ A2 N2

[2 marks]

b.i.

For writing M³ as M² × M or M × M² $\left( {{\text{or}}\left( {\begin{array}{*{20}{c}} 1&2 \\ 0&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1&4 \\ 0&1 \end{array}} \right)} \right)$ M1

M³ = $\left( {\begin{array}{*{20}{c}} {1 + 0}&{4 + 2} \\ {0 + 0}&{0 + 1} \end{array}} \right)$ A2

M³ = $\left( {\begin{array}{*{20}{c}} 1&6 \\ 0&1 \end{array}} \right)$ AG N0

[3 marks]

b.ii.

M⁴ = $\left( {\begin{array}{*{20}{c}} 1&8 \\ 0&1 \end{array}} \right)$ A1 N1

[1 mark]

c.i.

T₄ = $\left( {\begin{array}{*{20}{c}} 1&2 \\ 0&1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1&4 \\ 0&1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1&6 \\ 0&1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1&8 \\ 0&1 \end{array}} \right)$ (M1)

= $\left( {\begin{array}{*{20}{c}} 4&{20} \\ 0&4 \end{array}} \right)$ A1A1 N3

[3 marks]

c.ii.

T₁₀₀ = $\left( {\begin{array}{*{20}{c}} 1&2 \\ 0&1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1&4 \\ 0&1 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} 1&{200} \\ 0&1 \end{array}} \right)$ (M1)

$= \left( {\begin{array}{*{20}{c}} {100}&{10100} \\ 0&{100} \end{array}} \right)$ A1A1 N3

[3 marks]

Examiners’ report

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a.i.

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a.ii.

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b.i.

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b.ii.

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c.i.

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c.ii.

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