(c) Project Maths Development Team 2011

To prove by induction that (1 + x)n ≥ 1 + nx for x > -1, n N

Next

Note: x>-1. Hence (1+x)>0.

(c) Project Maths Development Team 2011

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

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 + kx2

Since (1 + x)k+1 ≥ 1 + kx + x + kx2 (k > 0 then kx2 ≥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

Prove true for n = k + 1