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 \(0 < a < 1\).

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

OR

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]