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content/post/exam-papers/2025高一竞赛/answer.md
@@ -182,7 +182,7 @@ $\therefore a \ge 0$
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当 $x \in \mathbb{N^*}$ 时,
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-- $f(2) > f(1) + f(1) = 2x > 2(x-1)$
+- $f(2) > f(1) + f(1) = 2 \cdot 2 > 2(2 - 1)$
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- $\forall n \in \mathbb{N^*}$, 若 $f(n) > 2n > 2(n-1)$, 则 $f(n+1) > f(n) + f(1) = f(n) + 2 > 2(n + 1) > 2n$
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$\therefore \forall x \in \mathbb{N^*}, f(x) > 2x > 2(x-1)$
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