If we use Mathematical Induction to prove that 3n≥2n[math]3^n \ge 2 n for all positive integers n[math]n, then the inductive step will be one of the following.

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This question asks about the inductive step used to prove 3^n ≥ 2n for all positive integers n. I'll evaluate each option in turn to see which one aligns with the typical induction pattern.

Option 1: Assuming that 3^k ≥ 2k, prove 3^{k-1} ≥ 2(k-1).
- Analysis: An induction step should move from n = k to n = k+1 (or from k to k+1 in the standard forward direction). Proving something about 3^{k-1} in terms of k-1 would move in the opposite direction, essentially stepping backwards in n. This does not establish the statement for n = k+1 based on the hypothesis for n = k, so this option is not a valid inductive step. It misaligns the exponent and the linear term with the forward progression of n.

Option 2: Assuming that 3^k ≥ 2k, prove 3^{k+1} ≥ 2(k+1).
- Analysis: This is the canonical inductive step. We assume the statement holds at n = k and show it holds at n = k+1 by deriving 3^{k+1} ≥ 2(k+1) from 3^k ≥ 2k, typically by multiplying both sides by 3 and then comparing to 2k+2, using simple algebra or inequalities. This directly advances n by 1 and preserves the required form, making it the correct inductive step.

Option 3: Assuming that 3^k ≥ 2k, prove 3^{k}+1 ≥ 2k+1.
- Analysis: The expression 3^{k}+1 is not the next term in the exponential sequence; the next term should be 3^{k+1}. This option attempts to tweak the left-hand side without properly increasing the exponent, so it does not establish the needed inductive progression from k to k+1 for the inequality 3^n ≥ 2n. Therefore, this is not the valid inductive step.

Option 4: Prove 3 ≥ 2.
- Analysis: While 3 ≥ 2 is true, it has nothing to do with the induction step for the statement involving 3^n and 2n for all positive integers n. Induction requires a relation between n = k and n = k+1, not a standalone truth about constants. This option fails to connect to the inductive framework and is irrelevant to proving the general inequality.

In summary, option 2 matches the standard inductive pattern by deducing the n = k+1 case from the n = k hypothesis, while the other options either move the index in the wrong direction, alter the expression improperly, or disregard the inductive structure entirely.

MATH 100 (LEC ER1 Winter 2025) Pre-class quiz 04. Due January 15 at 11:00 pm

If we use Mathematical Induction to prove that 3n≥2n[math]3^n \ge 2 n for all positive integers n[math]n, then the inductive step will be one of the following.

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