1 1 ∗ 2 + 1 3 ∗ 4 + 1 5 ∗ 6 + ⋯ + 1 ( 2 n − 1 ) 2 n = 1 n + 1 + 1 n + 2 + ⋯ + 1 2 n {\displaystyle {\frac {1}{1*2}}+{\frac {1}{3*4}}+{\frac {1}{5*6}}+\cdots +{\frac {1}{(2n-1)2n}}={\frac {1}{n+1}}+{\frac {1}{n+2}}+\cdots +{\frac {1}{2n}}}
L : 1 ( 2 n − 1 ) 2 n = 1 2 R : 1 2 n = 1 2 = 2 1 2 = 1 2 {\displaystyle {\begin{aligned}&L:{\frac {1}{(2n-1)2n}}={\frac {1}{2}}\\&R:{\frac {1}{2n}}={\frac {1}{2}}=2\\&{\frac {1}{2}}={\frac {1}{2}}\\\end{aligned}}}
1 1 ∗ 2 + 1 3 ∗ 4 + 1 5 ∗ 6 + ⋯ + 1 ( 2 k − 1 ) 2 k = 1 k + 1 + 1 k + 2 + ⋯ + 1 2 k {\displaystyle {\frac {1}{1*2}}+{\frac {1}{3*4}}+{\frac {1}{5*6}}+\cdots +{\frac {1}{(2k-1)2k}}={\frac {1}{k+1}}+{\frac {1}{k+2}}+\cdots +{\frac {1}{2k}}}
1 1 ∗ 2 + 1 3 ∗ 4 + 1 5 ∗ 6 + ⋯ + 1 ( 2 k − 1 ) 2 k ⏟ = 1 k + 1 + 1 k + 2 + ⋯ + 1 2 k + 1 ( 2 k + 1 ) 2 ( k + 1 ) = 1 k + 2 + 1 k + 3 ⋯ + 1 2 k + 1 2 k + 1 + 1 2 k + 2 1 k + 1 + 1 k + 2 + ⋯ + 1 2 k + 1 ( 2 k + 1 ) 2 ( k + 1 ) = 1 k + 2 + 1 k + 3 ⋯ + 1 2 k + 1 2 k + 1 + 1 2 k + 2 / − 1 k + 2 + 1 k + 3 ⋯ + 1 2 k 1 k + 1 + 1 ( 2 k + 1 ) 2 ( k + 1 ) = 1 2 k + 1 + 1 2 k + 2 / ∗ 2 ( 2 k + 1 ) 2 ( 2 k + 1 ) + 1 = 2 ( k + 1 ) + ( 2 k + 1 ) 4 k + 2 + 1 = 2 k + 2 + 24 + 1 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {1}{1*2}}+{\frac {1}{3*4}}+{\frac {1}{5*6}}+\cdots +{\frac {1}{(2k-1)2k}}} _{={\frac {1}{k+1}}+{\frac {1}{k+2}}+\cdots +{\frac {1}{2k}}}+{\frac {1}{(2k+1)2(k+1)}}={\frac {1}{k+2}}+{\frac {1}{k+3}}\cdots +{\frac {1}{2k}}+{\frac {1}{2k+1}}+{\frac {1}{2k+2}}\\&{\frac {1}{k+1}}+{\color {blue}{\frac {1}{k+2}}+\cdots +{\frac {1}{2k}}}+{\frac {1}{(2k+1)2(k+1)}}={\color {blue}{\frac {1}{k+2}}+{\frac {1}{k+3}}\cdots +{\frac {1}{2k}}}+{\frac {1}{2k+1}}+{\frac {1}{2k+2}}/-{\frac {1}{k+2}}+{\frac {1}{k+3}}\cdots +{\frac {1}{2k}}\\&{\frac {1}{k+1}}+{\frac {1}{(2k+1)2(k+1)}}={\frac {1}{2k+1}}+{\frac {1}{2k+2}}/*2(2k+1)\\&2(2k+1)+1=2(k+1)+(2k+1)\\&4k+2+1=2k+2+24+1\\&0=0\end{aligned}}}
האינדוקציה נכונה על פי 3 שלבי האינדוקציה.
טבעי
