3 2 − 5 6 + 7 12 − 9 20 + ⋯ + ( − 1 ) n − 1 ( 2 n + 1 ) n ( n + 1 ) = 1 + ( − 1 ) n + 1 n + 1 {\displaystyle {\frac {3}{2}}-{\frac {5}{6}}+{\frac {7}{12}}-{\frac {9}{20}}+\cdots +{\frac {(-1)^{n-1}(2n+1)}{n(n+1)}}=1+{\frac {(-1)^{n+1}}{n+1}}}
L : − ( − 1 ) n − 1 ( 2 n + 1 ) n ( n + 1 ) = ( − 1 ) 0 ( 2 + 1 ) 1 ∗ 2 = 1 ∗ 3 2 = 3 2 = 1.5 R : 1 + ( − 1 ) n + 1 n + 1 = 1 + ( − 1 ) n + 1 2 = 1.5 1.5 = 1.5 {\displaystyle {\begin{aligned}&L:-{\frac {(-1)^{n-1}(2n+1)}{n(n+1)}}={\frac {(-1)^{0}(2+1)}{1*2}}={\frac {1*3}{2}}={\frac {3}{2}}=1.5\\&R:1+{\frac {(-1)^{n+1}}{n+1}}=1+{\frac {(-1)^{n+1}}{2}}=1.5\\&1.5=1.5\\\end{aligned}}}
3 2 − 5 6 + 7 12 − 9 20 + ⋯ + ( − 1 ) k − 1 ( 2 k + 1 ) k ( k + 1 ) = 1 + ( − 1 ) k + 1 k + 1 {\displaystyle {\frac {3}{2}}-{\frac {5}{6}}+{\frac {7}{12}}-{\frac {9}{20}}+\cdots +{\frac {(-1)^{k-1}(2k+1)}{k(k+1)}}=1+{\frac {(-1)^{k+1}}{k+1}}}
3 2 − 5 6 + 7 12 − 9 20 + ⋯ + ( − 1 ) k − 1 ( 2 k + 1 ) k ( k + 1 ) ⏟ = 1 + ( − 1 ) k + 1 k + 1 + ( − 1 ) k ( 2 k + 3 ) ( k + 1 ) ( k + 2 ) = 1 + ( − 1 ) k + 2 k + 2 1 + ( − 1 ) k + 1 k + 1 + ( − 1 ) k ( 2 k + 3 ) ( k + 1 ) ( k + 2 ) = 1 + ( − 1 ) k + 2 k + 2 ( k + 2 ) ( − 1 ) k + 1 + ( − 1 ) k ( 2 k + 3 ) = ( k + 1 ) ( − 1 ) k + 2 ( k + 2 ) ( − 1 ) k + 1 − ( − 1 ) k + 1 ( 2 k + 3 ) = − ( k + 1 ) ( − 1 ) k + 1 / : ( − 1 ) k + 1 k + 2 − 2 k − 3 = − k − 1 − k − 1 = − k − 1 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {3}{2}}-{\frac {5}{6}}+{\frac {7}{12}}-{\frac {9}{20}}+\cdots +{\frac {(-1)^{k-1}(2k+1)}{k(k+1)}}} _{=1+{\frac {(-1)^{k+1}}{k+1}}}+{\frac {(-1)^{k}(2k+3)}{(k+1)(k+2)}}=1+{\frac {(-1)^{k+2}}{k+2}}\\&1+{\frac {(-1)^{k+1}}{k+1}}+{\frac {(-1)^{k}(2k+3)}{(k+1)(k+2)}}=1+{\frac {(-1)^{k+2}}{k+2}}\\&(k+2)(-1)^{k+1}+(-1)^{k}(2k+3)=(k+1)(-1)^{k+2}\\&(k+2)(-1)^{k+1}-(-1)^{k+1}(2k+3)=-(k+1)(-1)^{k+1}/:(-1)^{k+1}\\&k+2-2k-3=-k-1\\&-k-1=-k-1\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
