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lucifer
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chore: 优化文案
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problems/1227.airplane-seat-assignment-probability.md

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@@ -173,14 +173,16 @@ f(n)
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f(n-1) = 1/(n-1) * (f(n-2) + f(n-3) + ... + f(1))
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```
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我们将等式 1 和等式 2 两边分别同时乘以 n 和 n - 1
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我们将等式 1 和等式 2 两边分别同时乘以 n 和 n - 1
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```
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n * f(n) = f(n-1) + f(n-2) + f(n-3) + ... + f(1)
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(n-1) * f(n-1) = f(n-2) + f(n-3) + ... + f(1)
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```
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我们将两者相减:
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之后我们使用错位相减技巧可以将等式进一步花间。
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具体来说我们可以将两者相减:
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```
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n * f(n) - (n-1)*f(n-1) = f(n-1)

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