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