1 1 ∗ 5 + 1 5 ∗ 9 + 1 9 ∗ 13 + ⋯ + 1 ( 4 n − 3 ) ( 4 n + 1 ) = n 4 n + 1 {\displaystyle {\frac {1}{1*5}}+{\frac {1}{5*9}}+{\frac {1}{9*13}}+\cdots +{\frac {1}{(4n-3)(4n+1)}}={\frac {n}{4n+1}}}
L : 1 ( 4 n − 3 ) ( 4 n + 1 = 1 ( 4 − 3 ) ( 4 + 1 ) = 1 5 R : n 4 n + 1 = 1 4 + 1 = 1 5 1 5 = 1 5 √ {\displaystyle {\begin{aligned}&L:{\frac {1}{(4n-3)(4n+1}}={\frac {1}{(4-3)(4+1)}}={\frac {1}{5}}\\&R:{\frac {n}{4n+1}}={\frac {1}{4+1}}={\frac {1}{5}}\\&{\frac {1}{5}}={\frac {1}{5}}\surd \\\end{aligned}}}
1 1 ∗ 5 + 1 5 ∗ 9 + 1 9 ∗ 13 + ⋯ + 1 ( 4 k − 3 ) ( 4 k + 1 ) = k 4 k + 1 {\displaystyle {\frac {1}{1*5}}+{\frac {1}{5*9}}+{\frac {1}{9*13}}+\cdots +{\frac {1}{(4k-3)(4k+1)}}={\frac {k}{4k+1}}}
1 1 ∗ 5 + 1 5 ∗ 9 + 1 9 ∗ 13 + ⋯ + 1 ( 4 k − 3 ) ( 4 k + 1 ) ⏟ = k 4 k + 1 + 1 ( 4 k + 1 ) ( 4 k + 5 ) = ( k + 1 ) 4 k + 5 k 4 k + 1 + 1 ( 4 k + 1 ) ( 4 k + 5 ) = ( k + 1 ) 4 k + 5 k ( 4 k + 5 ) + 1 = ( k + 1 ) ( 4 k + 1 ) 4 k 2 + 5 k + 1 = 4 k 2 + 5 k + 1 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {1}{1*5}}+{\frac {1}{5*9}}+{\frac {1}{9*13}}+\cdots +{\frac {1}{(4k-3)(4k+1)}}} _{={\frac {k}{4k+1}}}+{\frac {1}{(4k+1)(4k+5)}}={\frac {(k+1)}{4k+5}}\\&{\frac {k}{4k+1}}+{\frac {1}{(4k+1)(4k+5)}}={\frac {(k+1)}{4k+5}}\\&k(4k+5)+1=(k+1)(4k+1)\\&4k^{2}+5k+1=4k^{2}+5k+1\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
