5 4 − 11 28 + 17 70 − ⋯ + ( − 1 ) n − 1 ( 6 n − 1 ) 3 n − 2 ) ( 3 n + 1 ) = 1 + ( − 1 ) n + 1 3 n + 1 {\displaystyle {\frac {5}{4}}-{\frac {11}{28}}+{\frac {17}{70}}-\cdots +{\frac {(-1)^{n-1}(6n-1)}{3n-2)(3n+1)}}=1+{\frac {(-1)^{n+1}}{3n+1}}}
L : ( − 1 ) n − 1 ( 6 n − 1 ) 3 n − 2 ) ( 3 n + 1 ) = ( − 1 ) 0 ( 6 − 1 ) ( 3 − 2 ) ( 3 + 1 ) = 5 4 = 1.25 R : 1 + ( − 1 ) n + 1 3 n + 1 = 1 + ( − 1 ) 2 3 + 1 = 1.25 1.25 = 1.25 {\displaystyle {\begin{aligned}&L:{\frac {(-1)^{n-1}(6n-1)}{3n-2)(3n+1)}}={\frac {(-1)^{0}(6-1)}{(3-2)(3+1)}}={\frac {5}{4}}=1.25\\&R:1+{\frac {(-1)^{n+1}}{3n+1}}=1+{\frac {(-1)^{2}}{3+1}}=1.25\\&1.25=1.25\\\end{aligned}}}
5 4 − 11 28 + 17 70 − ⋯ + ( − 1 ) k − 1 ( 6 k − 1 ) 3 k − 2 ) ( 3 k + 1 ) = 1 + ( − 1 ) k + 1 3 k + 1 {\displaystyle {\frac {5}{4}}-{\frac {11}{28}}+{\frac {17}{70}}-\cdots +{\frac {(-1)^{k-1}(6k-1)}{3k-2)(3k+1)}}=1+{\frac {(-1)^{k+1}}{3k+1}}}
5 4 − 11 28 + 17 70 − ⋯ + ( − 1 ) k − 1 ( 6 k − 1 ) 3 k − 2 ) ( 3 k + 1 ) ⏟ = 1 + ( − 1 ) k + 1 3 k + 1 + ( − 1 ) k ( 6 k + 5 ) ( 3 k + 1 ) ( 3 k + 4 ) = 1 + ( − 1 ) k + 2 3 k + 4 1 + ( − 1 ) k + 1 3 k + 1 + ( − 1 ) k ( 6 k + 5 ) ( 3 k + 1 ) ( 3 k + 4 ) = 1 + ( − 1 ) k + 2 3 k + 4 ( 3 k + 4 ) ( − 1 ) k + 1 + ( − 1 ) k ( 6 k + 5 ) = ( 3 k + 1 ) ( − 1 ) k + 2 ( 3 k + 4 ) ( − 1 ) k + 1 − ( − 1 ) k + 1 ( 6 k + 5 ) = − ( 3 k + 1 ) ( − 1 ) k + 1 3 k + 4 − 6 k − 5 = − 3 k − 1 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {5}{4}}-{\frac {11}{28}}+{\frac {17}{70}}-\cdots +{\frac {(-1)^{k-1}(6k-1)}{3k-2)(3k+1)}}} _{=1+{\frac {(-1)^{k+1}}{3k+1}}}+{\frac {(-1)^{k}(6k+5)}{(3k+1)(3k+4)}}=1+{\frac {(-1)^{k+2}}{3k+4}}\\&1+{\frac {(-1)^{k+1}}{3k+1}}+{\frac {(-1)^{k}(6k+5)}{(3k+1)(3k+4)}}=1+{\frac {(-1)^{k+2}}{3k+4}}\\&(3k+4)(-1)^{k+1}+(-1)^{k}(6k+5)=(3k+1)(-1)^{k+2}\\&(3k+4)(-1)^{k+1}-(-1)^{k+1}(6k+5)=-(3k+1)(-1)^{k+1}\\&3k+4-6k-5=-3k-1\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.