Question

Use mathematical induction to justify that if 1+x>0, then for all n € Ñ, (1+x)^n>1+nx

Answer 1

Not true for n=0,1, but check for n>1

case n=1: (1+x)^2 = 1+2x+x^2 > 1+2x
assume true for n=k, and check n=k+1:
(1+x)^(k+1) = (1+x)^k * (1+x) > (1+kx)(1+x) = (1+kx) + x + kx^2 = 1 + (k+1)x
QED