Proof by Induction (n=k+1 steps difficulty)

Question

Use mathematical induction to prove that $( 1 − a ) ^ n > 1 − na$ for $\{ n : n ∈ Z + , n ⩾ 2 \}$ where $0 < a < 1$ .

I understand the first steps but I don't understand the final ones (the $n = k+1$ steps). Please help me out if possible.

Answer 1

Using induction hypothesis we have for $k+1$: $(1-a)^{k+1}=(1-a)^k \cdot (1-a)> (1-ka)(1-a)=1-a-ka+ka^2>1-a(k+1)$

Answer 2

Let $P(n) = (1- a)^n, 0<a<1$

For $n = 2, P(2) = (1-a)^2 = 1- 2a + a^2 > 1 -2a$

Thus P(2) is true.

Let's assume P(n) is true for some n = k($\epsilon$Z)

So,$$P(k) = (1 - a)^k > 1- ka ---(i)$$ Now we have to prove P(k+1) is true. $$i.e, P(k+1) = (1- a)^{k+1} > 1-(k+1)a ---(ii)$$ From inequality(i),$$LHS = P(k+1) = (1-a)^{k+1} = (1-a)^k.(1-a) > (1-ka).(1-a)$$ $$LHS>(1 - a - ka - a^2)$$ $$ LHS>(1-a(k+1) -a^2)$$ As $ 0<a<1, LHS>(1-(k+1)a).$ Hence Proved!!!