Question

Use mathematical induction to prove for any n≥1 and n∈N, we have 1/2ⁿ < 2/2+n

Answer 1

To prove the inequality using mathematical induction, we need to prove the base case and the inductive step. Assuming the inequality is true for n = k, we need to prove it for n = k + 1. Therefore, the inequality is true for all n ≥ 1 and n ∈ N.

Step-by-step explanation:

To prove the inequality using mathematical induction, we need to prove the base case and the inductive step.

Base case:

When n = 1, we have
`1/2^1 = 1/2 < 2/2+1 = 2/3,`

Inductive step:

Assuming the inequality is true for n = k, we need to prove it for
`n = k + 1.`

So, we need to prove that
`1/2^(k+1) < 2/2+(k+1)`< 2/2+k, which is our inductive hypothesis.

Multiplying both sides of the inductive hypothesis by 1/2, we get 1/2^(k+1) < 1/2(2/2+k).

Simplifying the right side, we have 1/2^(k+1) < 1/(2+k), which is true since 1/2+k is larger than 1 and therefore 1/(2+k) is smaller than 1.

Therefore, the inequality is true for all
`n ≥ 1`∈ N.