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.

Question

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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.

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.单项选择题

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