1 1 ∗ 3 ∗ 5 + 2 3 ∗ 5 ∗ 7 + ⋯ + n ( 2 n − 1 ) ( 2 n + 1 ) ( 2 n + 3 ) = n ( n + 1 ) 2 ( 2 n + 1 ) ( 2 n + 3 ) {\displaystyle {\frac {1}{1*3*5}}+{\frac {2}{3*5*7}}+\cdots +{\frac {n}{(2n-1)(2n+1)(2n+3)}}={\frac {n(n+1)}{2(2n+1)(2n+3)}}}
L : n ( 2 n − 1 ) ( 2 n + 1 ) ( 2 n + 3 ) = 1 1 ∗ 3 ∗ 5 = 1 15 R : n ( n + 1 ) 2 ( 2 n + 1 ) ( 2 n + 3 ) = 2 2 ∗ 3 ∗ 5 = 1 15 1 5 = 1 5 √ {\displaystyle {\begin{aligned}&L:{\frac {n}{(2n-1)(2n+1)(2n+3)}}={\frac {1}{1*3*5}}={\frac {1}{15}}\\&R:{\frac {n(n+1)}{2(2n+1)(2n+3)}}={\frac {2}{2*3*5}}={\frac {1}{15}}\\&{\frac {1}{5}}={\frac {1}{5}}\surd \\\end{aligned}}}
1 1 ∗ 3 ∗ 5 + 2 3 ∗ 5 ∗ 7 + ⋯ + k ( 2 k − 1 ) ( 2 k + 1 ) ( 2 k + 3 ) = k ( k + 1 ) 2 ( 2 k + 1 ) ( 2 k + 3 ) {\displaystyle {\frac {1}{1*3*5}}+{\frac {2}{3*5*7}}+\cdots +{\frac {k}{(2k-1)(2k+1)(2k+3)}}={\frac {k(k+1)}{2(2k+1)(2k+3)}}}
1 1 ∗ 3 ∗ 5 + 2 3 ∗ 5 ∗ 7 + ⋯ + k ( 2 k − 1 ) ( 2 k + 1 ) ( 2 k + 3 ) ⏟ = k ( k + 1 ) 2 ( 2 k + 1 ) ( 2 k + 3 ) + k + 1 ( 2 k + 1 ) ( 2 k + 3 ) ( 2 k + 5 ) = ( k + 1 ) ( k + 2 ) 2 ( 2 k + 3 ) ( 2 k + 5 ) k ( k + 1 ) ( 2 k + 5 ) + 2 ( k + 1 ) = ( k + 1 ) ( k + 2 ) ( 2 k + 1 ) k ( 2 k + 5 ) + 2 = ( k + 2 ) ( 2 k + 1 ) 2 k 2 + 5 k + 2 = 2 k 2 + k + 4 k + 2 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {1}{1*3*5}}+{\frac {2}{3*5*7}}+\cdots +{\frac {k}{(2k-1)(2k+1)(2k+3)}}} _{={\frac {k(k+1)}{2(2k+1)(2k+3)}}}+{\frac {k+1}{(2k+1)(2k+3)(2k+5)}}={\frac {(k+1)(k+2)}{2(2k+3)(2k+5)}}\\&k(k+1)(2k+5)+2(k+1)=(k+1)(k+2)(2k+1)\\&k(2k+5)+2=(k+2)(2k+1)\\&2k^{2}+5k+2=2k^{2}+k+4k+2\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
