1 n ( n + 1 ) + 1 ( n + 1 ) ( n + 2 ) + ⋯ + 1 ( 2 n − 1 ) 2 n = 1 2 n {\displaystyle {\frac {1}{n(n+1)}}+{\frac {1}{(n+1)(n+2)}}+\cdots +{\frac {1}{(2n-1)2n}}={\frac {1}{2n}}}
L : 1 ( 2 n − 1 ) 2 n = 1 ( 2 − 1 ) 2 = 1 2 → 1 1 ( 1 + 1 ) = 1 2 R : 1 2 n = 1 2 0.5 = 0.5 {\displaystyle {\begin{aligned}&L:{\frac {1}{(2n-1)2n}}={\frac {1}{(2-1)2}}={\frac {1}{2}}\rightarrow {\frac {1}{1(1+1)}}={\frac {1}{2}}\\&R:{\frac {1}{2n}}={\frac {1}{2}}\\&0.5=0.5\\\end{aligned}}}
1 k ( k + 1 ) + 1 ( k + 1 ) ( k + 2 ) + ⋯ + 1 ( 2 k − 1 ) 2 k = 1 2 k {\displaystyle {\frac {1}{k(k+1)}}+{\frac {1}{(k+1)(k+2)}}+\cdots +{\frac {1}{(2k-1)2k}}={\frac {1}{2k}}}
1 ( k + 1 ) ( k + 2 ) + 1 ( k + 1 ) ( k + 3 ) + ⋯ + 1 ( 2 k − 1 ) 2 k ⏟ = 1 2 k − 1 k ( k + 1 ) + 1 ( 2 k ) ( 2 k + 1 ) + 2 k + 1 2 k + 2 = 1 2 k + 2 1 2 k − 1 k ( k + 1 ) + 1 ( 2 k ) ( 2 k + 1 ) + 2 k + 1 2 k + 2 = 1 2 k + 2 1 2 k − 1 k ( k + 1 ) + 1 2 k ( 2 k + 1 ) + 1 2 ( 2 k + 1 ) ( k + 1 ) = 1 2 ( k + 1 ) ( k + 1 ) ( 2 k + 1 ) − 2 ( 2 k + 1 ) + k + 1 + k = k ( 2 k + 1 ) 2 k 2 + k + 2 k + 1 − 4 k − 2 + 2 k + 1 = 2 k 2 + k 2 k 2 + k = 2 k 2 + k 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {1}{(k+1)(k+2)}}+{\frac {1}{(k+1)(k+3)}}+\cdots +{\frac {1}{(2k-1)2k}}} _{={\frac {1}{2k}}-{\frac {1}{k(k+1)}}}+{\frac {1}{(2k)(2k+1)}}+{\frac {2k+1}{2k+2}}={\frac {1}{2k+2}}\\&{\frac {1}{2k}}-{\frac {1}{k(k+1)}}+{\frac {1}{(2k)(2k+1)}}+{\frac {2k+1}{2k+2}}={\frac {1}{2k+2}}\\&{\frac {1}{2k}}-{\frac {1}{k(k+1)}}+{\frac {1}{2k(2k+1)}}+{\frac {1}{2(2k+1)(k+1)}}={\frac {1}{2(k+1)}}\\&(k+1)(2k+1)-2(2k+1)+k+1+k=k(2k+1)\\&2k^{2}+k+2k+1-4k-2+2k+1=2k^{2}+k\\&2k^{2}+k=2k^{2}+k\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
