1. To prove by induction that (1 + x) n ≥ 1 + nx for x > -1, n  N Next (c) Project Maths Development Team 2011 Note: x>-1. Hence (1+x)>0.

2. Prove: (1 + x) n ≥ 1 + nx for n=1 For n = 1 (1 + x) 1 = 1 + x True for n = 1 To prove by induction that (1 + x) n ≥ 1 + nx for x > -1, n  N Next (c) Project Maths Development Team 2011

3. Assume true for n = k. Therefore (1 + x) k ≥ 1 + kx Multiply each side by 1 + x (1 + x)(1 + x) k ≥ (1 + x)(1 + kx) (1 + x) k+1 ≥ 1 + kx + x + kx 2 Since (1 + x) k+1 ≥ 1 + kx + x + kx 2 (k > 0 then kx 2 ≥0 for all x) Therefore (1 + x) k+1 ≥ 1 + kx + x (1 + x) k+1 ≥ 1 + (k+1)x If true for n = k this implies it is true for n = k + 1. It is true n = 1. Hence by induction (1 + x) n ≥ 1 + nx x > -1, n  N. To prove by induction that (1 + x) n ≥ 1 + nx for x > -1, n  N (c) Project Maths Development Team 2011 Prove true for n = k + 1