# Prove by the method of Mathematical induction that $(1-0.3)^n \geq 1-0.3n$ for all $n$ in set of positive integers

Here is what I have so far

Basis

For $n = 0 (1-0.3)^0 \geq 1-0.3(0)$ checks

For $n = 1 (1-0.3)^1 \geq 1-0.3(k$) checks

I.H. $(1-0.3)^k \geq 1-0.3(k)$ for all k in the set of positive integers (1)

We want to prove that $(1-0.3)^{k+1} \geq 1-0.3(k+1)$ (2)

To relate (1) to (2) we have to add $(1-0.3)^{k+1}$ to both sides of (1).

$(1-0.3)^k+(1-0.3)^{k+1} \geq 1-0.3(k) + (1-0.3)^{k+1}$.

Here is where I am stuck. I need help as to where to go from here.

• Hint : Try binomial expansion of (1-0.3)^k+1 in equation 2 . – Lakshya Gupta Apr 24 '17 at 23:52
• "To relate (1) to (2) we have to add (1−0.3)k+1 to both sides of (1)." Why? – fleablood Apr 25 '17 at 0:13

I don't see why you have to add $(1-0.3)^{k+1}$.