Answer to Question #189930 in Algebra for OhMoe

Question

We do a proof by induction to show the following inequality is true for all n∈N: 4ⁿ ≥ 4n

Let P(n) represent the statement: 4ⁿ ≥ 4n

For n=1: LHS = 4¹ = 4 and RHS = 4(1) = 4

Since 4≥4, P(1) is true.

Assume true for n=k: P(k)true means 4^k ≥ 4k

Show true for n=k+1 assuming P(k):

Provide the correct proof.

Answer 1

Given statement is-

$P(n):4^n\ge 4n ,\forall n\in N$

For $n=1, LHS= 4^1=4$

$RHS=4(1)=4=LHS$

Hence P(n) is true for n=1.

Let us assume that p(k) is true for some positive integer k.

$P(k):4^k\ge 4k~~~~~~-(1)$

We want to prove p(k+1) is true when p(k) is true-

$P(k+1): LHS= 4^{k+1}=4^k.4$

$RHS=4(k+1)=4k+4$

As $4^k\ge 4\Rightarrow 4^k.4\ge 4k+4$

So $P(k+1)$ is true.

So given statement is true for all $n\in N.$