( 1 − 1 ( n + 1 ) 2 ) ( 1 − 1 ( n + 2 ) 2 ) ( 1 − 1 ( n + 3 ) 2 ) + ⋯ + ( 1 − 1 ( 2 n ) 2 ) = 2 n + 1 2 n + 2 {\displaystyle (1-{\frac {1}{(n+1)^{2}}})(1-{\frac {1}{(n+2)^{2}}})(1-{\frac {1}{(n+3)^{2}}})+\cdots +(1-{\frac {1}{(2n)^{2}}})={\frac {2n+1}{2n+2}}}
L : ( 1 − 1 ( 2 n ) 2 ) = 1 − 1 4 → 1 − 1 ( 1 + 1 ) 2 = 1 − 0.25 = 3 4 R : 2 n + 1 2 n + 2 = 2 + 1 2 + 2 = 0.75 0.75 = 0.75 {\displaystyle {\begin{aligned}&L:(1-{\frac {1}{(2n)^{2}}})=1-{\frac {1}{4}}\rightarrow 1-{\frac {1}{(1+1)^{2}}}=1-0.25={\frac {3}{4}}\\&R:{\frac {2n+1}{2n+2}}={\frac {2+1}{2+2}}=0.75\\&0.75=0.75\\\end{aligned}}}
( 1 − 1 ( k + 1 ) 2 ) ( 1 − 1 ( k + 2 ) 2 ) ( 1 − 1 ( k + 3 ) 2 ) + ⋯ + ( 1 − 1 ( 2 k ) 2 ) = 2 k + 1 2 k + 2 {\displaystyle (1-{\frac {1}{(k+1)^{2}}})(1-{\frac {1}{(k+2)^{2}}})(1-{\frac {1}{(k+3)^{2}}})+\cdots +(1-{\frac {1}{(2k)^{2}}})={\frac {2k+1}{2k+2}}}
( k + 1 ) 2 ) ( 1 − 1 ( k + 2 ) 2 ) ( 1 − 1 ( k + 3 ) 2 ) + ⋯ + ( 1 − 1 ( 2 k ) 2 ) ⏟ = 2 k + 1 2 k + 2 ( 1 − 1 k + 1 2 ) ( 1 − 1 ( 2 k + 1 ) 2 ) ( 1 − 1 ( 2 k + 2 ) 2 ) = 2 k + 3 2 k + 4 2 k + 1 2 k + 2 ( 1 − 1 k + 1 2 ) ( 1 − 1 ( 2 k + 1 ) 2 ) ( 1 − 1 ( 2 k + 2 ) 2 ) = 2 k + 3 2 k + 4 2 k + 1 2 k + 2 ( ( k + 1 ) 2 − 1 k + 1 2 ) ( 1 − 1 ( 2 k + 1 ) 2 ) ( 1 − 1 ( 2 k + 2 ) 2 ) = 2 k + 3 2 k + 4 2 k + 1 2 k + 2 ∗ ( k + 1 ) 2 ( k + 1 ) 2 − 1 ∗ ( 2 k + 1 ) 2 − 1 ( 2 k + 1 ) 2 ∗ ( 2 k + 2 ) 2 − 1 ( 2 k + 2 ) 2 = 2 k + 3 2 k + 4 2 k + 1 2 k + 2 ∗ ( k + 1 ) 2 k 2 + 2 k ∗ 4 k 2 + 4 k ( 2 k + 1 ) 2 ∗ 4 k 2 + 8 k + 3 2 ( k + 1 ) 2 = 2 k + 3 2 ( k + 2 ) 2 k + 1 2 ( k + 1 ) ∗ ( k + 1 ) 2 k ( k + 2 ) ∗ 4 k ( k + 1 ) ( 2 k + 1 ) 2 ∗ 4 k 2 + 8 k + 3 2 ( k + 1 ) 2 = 2 k + 3 2 ( k + 2 ) 1 ( 2 k + 3 ) 2 ( k + 2 ) = 2 k + 3 2 ( k + 2 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{(k+1)^{2}})(1-{\frac {1}{(k+2)^{2}}})(1-{\frac {1}{(k+3)^{2}}})+\cdots +(1-{\frac {1}{(2k)^{2}}})} _{={\frac {\frac {2k+1}{2k+2}}{(1-{\frac {1}{k+1}}^{2})}}}(1-{\frac {1}{(2k+1)^{2}}})(1-{\frac {1}{(2k+2)^{2}}})={\frac {2k+3}{2k+4}}\\&{\frac {\frac {2k+1}{2k+2}}{(1-{\frac {1}{k+1}}^{2})}}(1-{\frac {1}{(2k+1)^{2}}})(1-{\frac {1}{(2k+2)^{2}}})={\frac {2k+3}{2k+4}}\\&{\frac {\frac {2k+1}{2k+2}}{({\frac {(k+1)^{2}-1}{k+1}}^{2})}}(1-{\frac {1}{(2k+1)^{2}}})(1-{\frac {1}{(2k+2)^{2}}})={\frac {2k+3}{2k+4}}\\&{\frac {2k+1}{2k+2}}*{\frac {(k+1)^{2}}{(k+1)^{2}-1}}*{\frac {(2k+1)^{2}-1}{(2k+1)^{2}}}*{\frac {(2k+2)^{2}-1}{(2k+2)^{2}}}={\frac {2k+3}{2k+4}}\\&{\frac {2k+1}{2k+2}}*{\frac {(k+1)^{2}}{k^{2}+2k}}*{\frac {4k^{2}+4k}{(2k+1)^{2}}}*{\frac {4k^{2}+8k+3}{2(k+1)^{2}}}={\frac {2k+3}{2(k+2)}}\\&{\frac {2k+1}{2(k+1)}}*{\frac {(k+1)^{2}}{k(k+2)}}*{\frac {4k(k+1)}{(2k+1)^{2}}}*{\frac {4k^{2}+8k+3}{2(k+1)^{2}}}={\frac {2k+3}{2(k+2)}}\\&{\frac {1(2k+3)}{2(k+2)}}={\frac {2k+3}{2(k+2}}\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
