數學歸納法的通式

出自高材生

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若為一數學命題，則
1. $P\left( 1 \right)\And \left[ P\left( k \right)\Rightarrow P\left( k+1 \right) \right]\Rightarrow \left[ P(n),\forall n\in N \right]$
2. $P\left( 1 \right)\And P\left( 2 \right)\And \left[ P\left( k \right)\And P\left( k+1 \right)\Rightarrow P\left( k+2 \right) \right]\Rightarrow \left[ P(n),\forall n\in N \right]$
3. $P\left( 1 \right)\And P\left( 2 \right)\And P\left( 3 \right)\And \left[ P\left( k \right)\And P\left( k+1 \right)\And P\left( k+2 \right)\Rightarrow P\left( k+3 \right) \right]\Rightarrow \left[ P(n),\forall n\in N \right]$
4. $\vdots$
5. $P\left( 1 \right)\And P\left( 2 \right)\And \cdots \And P\left( i \right)\And \left[ P\left( k \right)\And P\left( k+1 \right)\And \cdots \And P\left( k+i-1 \right)\Rightarrow P\left( k+i \right) \right]\Rightarrow \left[ P(n),\forall n\in N \right]$

取自"http://www.come2pass.com/wiki/index.php/%E6%95%B8%E5%AD%B8%E6%AD%B8%E7%B4%8D%E6%B3%95%E7%9A%84%E9%80%9A%E5%BC%8F"

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