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