Use mathematical induction to prove that \((2n)! \ge {2^n}{(n!)^2},{\text{ }}n \in {\mathbb{Z}^ + }\).

let \({\text{P}}(n)\) be the proposition that \((2n)! \ge {2^n}{(n!)^2},{\text{ }}n \in {\mathbb{Z}^ + }\)

consider \({\text{P}}(1)\):

\(2! = 2\) and \({2^1}{(1!)^2} = 2\) so \({\text{P}}(1)\) is true     R1

assume \({\text{P}}(k)\) is true ie \((2k)! \ge {2^k}{(k!)^2},{\text{ }}n \in {\mathbb{Z}^ + }\)     M1

Note:     Do not award M1 for statements such as “let \(n = k\)”.

consider \({\text{P}}(k + 1)\):

\(\left( {2(k + 1)} \right)! = (2k + 2)(2k + 1)(2k)!\)     M1

\(\left( {2(k + 1)} \right)! \ge (2k + 2)(2k + 1){(k!)^2}{2^k}\)     A1

\( = 2(k + 1)(2k + 1){(k!)^2}{2^k}\)

\( > {2^{k + 1}}(k + 1)(k + 1){(k!)^2}\;\;\;{\text{since}}\;\;\;2k + 1 > k + 1\)     R1

\( = {2^{k + 1}}{\left( {(k + 1)!} \right)^2}\)     A1

\({\text{P}}(k + 1)\) is true whenever \({\text{P}}(k)\) is true and \({\text{P}}(1)\) is true, so \({\text{P}}(n)\) is true for \(n \in {\mathbb{Z}^ + }\)     R1

Note:     To obtain the final R1, four of the previous marks must have been awarded.

[7 marks]

An easy question, but many candidates exhibited discomfort and poor reasoning abilities. The difficulty for most was that the proposition was expressed in terms of an inequality. Hopefully, as most publishers of IB textbooks have realised, inequalities in such questions are within the syllabus.