18M.2.AHL.TZ2.H_6

pestleMathematicsAAHLPaper 218M· ahl-1-15-proof-by-induction-contradiction-counterexamplessource ↗

Use mathematical induction to prove that ${\left( {1 - a} \right)^n} > 1 - na$ for $\left\{ {n\,{\text{:}}\,n \in {\mathbb{Z}^ + },\,n \geqslant 2} \right\}$ where $0 < a < 1$ .

Markscheme / solution

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

Let ${{\text{P}}_n}$ be the statement: ${\left( {1 - a} \right)^n} > 1 - na$ for some ${n \in {\mathbb{Z}^ + },\,n \geqslant 2}$ where $0 < a < 1$ consider the case $n = 2{\text{:}}\,\,\,{\left( {1 - a} \right)^2} = 1 - 2a + {a^2}$ M1

$> 1 - 2a$ because ${a^2} < 0$ . Therefore ${{\text{P}}_2}$ is true R1

assume ${{\text{P}}_n}$ is true for some $n = k$

${\left( {1 - a} \right)^k} > 1 - ka$ M1

Note: Assumption of truth must be present. Following marks are not dependent on this M1.

EITHER

consider ${\left( {1 - a} \right)^{k + 1}} = \left( {1 - a} \right){\left( {1 - a} \right)^k}$ M1

$> 1 - \left( {k + 1} \right)a + k{a^2}$ A1

$> 1 - \left( {k + 1} \right)a \Rightarrow {{\text{P}}_{k + 1}}$ is true (as $k{a^2} > 0$ ) R1

multiply both sides by $\left( {1 - a} \right)$ (which is positive) M1

${\left( {1 - a} \right)^{k + 1}} > \left( {1 - ka} \right)\left( {1 - a} \right)$

${\left( {1 - a} \right)^{k + 1}} > 1 - \left( {k + 1} \right)a + k{a^2}$ A1

${\left( {1 - a} \right)^{k + 1}} > 1 - \left( {k + 1} \right)a \Rightarrow {{\text{P}}_{k + 1}}$ is true (as $k{a^2} > 0$ ) R1

THEN

${{\text{P}}_2}$ is true ${{\text{P}}_k}$ is true $\Rightarrow {{\text{P}}_{k + 1}}$ is true so ${{\text{P}}_n}$ true for all $n > 2$ (or equivalent) R1

Note: Only award the last R1 if at least four of the previous marks are gained including the A1.

[7 marks]

Examiners’ report

[N/A]