Final answer:
To prove by induction that (1−x)^n ≥ 1−nx, we verify the base case n=1. Then we assume the statement is true for k and prove it for k+1 by following the inductive step, showing the inequality holds for every natural number n ≥ 1 with 0 ≤ x ≤ 1.
Step-by-step explanation:
We are asked to prove by induction that (1−x)^n ≥1−nx for any natural number n ≥ 1 and 0 ≤ x ≤ 1.
Base Case (n=1):
For n=1, (1−x)^1 = 1−x is obviously equal to 1−(1)x, which satisfies our initial condition. Hence, the statement holds for n=1.
Inductive Step:
Assume that the statement (1−x)^k ≥ 1−kx is true for some natural number k. Our goal is to prove that the statement also holds for k+1, i.e., (1−x)^(k+1) ≥ 1−(k+1)x.
We start with the left-hand side of our assumption and multiply both sides by (1-x):
- (1−x)^k * (1−x) ≥ (1−kx)*(1−x)
- (1−x)^(k+1) ≥ 1−(k+1)x + kx^2
- (1−x)^(k+1) ≥ 1−(k+1)x
Because kx^2 ≥ 0 (since 0 ≤ x ≤ 1), we know that 1−(k+1)x + kx^2 ≥ 1−(k+1)x. Thus, (1−x)^(k+1) ≥ 1−(k+1)x, and the statement is true for k+1.