( 1 − 1 4 ) ( 1 − 1 9 ( 1 − 1 16 ) + ⋯ + ( 1 − 1 n 2 = n + 1 2 n {\displaystyle (1-{\frac {1}{4}})(1-{\frac {1}{9}}(1-{\frac {1}{16}})+\cdots +(1-{\frac {1}{n^{2}}}={\frac {n+1}{2n}}}
L : ( 1 − 1 n 2 = 1 − 1 4 = 3 4 R : n + 1 2 n = + 1 2 + 2 = 3 4 3 4 = 3 4 √ {\displaystyle {\begin{aligned}&L:(1-{\frac {1}{n^{2}}}=1-{\frac {1}{4}}={\frac {3}{4}}\\&R:{\frac {n+1}{2n}}={\frac {+1}{2+2}}={\frac {3}{4}}\\&{\frac {3}{4}}={\frac {3}{4}}\surd \\\end{aligned}}}
( 1 − 1 4 ) ( 1 − 1 9 ( 1 − 1 16 ) + ⋯ + ( 1 − 1 k 2 = k + 1 2 k ) {\displaystyle (1-{\frac {1}{4}})(1-{\frac {1}{9}}(1-{\frac {1}{16}})+\cdots +(1-{\frac {1}{k^{2}}}={\frac {k+1}{2k}})}
( 1 4 ) ( 1 − 1 9 ) ( 1 − 1 16 ) + ⋯ + ( 1 − 1 k 2 ) ⏟ = k + 1 2 k + ( 1 − 1 ( k + 1 ) 2 = k + 2 2 ( k + 1 ) ) k + 1 2 k + ( 1 − 1 ( k + 1 ) 2 ) = k + 2 2 ( k + 1 ) ( k + 1 ) ( k + 1 ) 2 − 1 2 k ∗ ( k + 1 ) = k + 2 2 ( k + 1 ) ( k + 1 ) ( k 2 + 2 k ) = k ( k + 1 ) ( k + 2 ) k ( k + 1 ) ( k + 2 ) = k ( k + 1 ) ( k + 2 ) 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {({\frac {1}{4}})(1-{\frac {1}{9}})(1-{\frac {1}{16}})+\cdots +(1-{\frac {1}{k^{2}}})} _{={\frac {k+1}{2k}}}+(1-{\frac {1}{(k+1)^{2}}}={\frac {k+2}{2(k+1)}})\\&{\frac {k+1}{2k}}+(1-{\frac {1}{(k+1)^{2}}})={\frac {k+2}{2(k+1)}}\\&{\frac {(k+1)(k+1)^{2}-1}{2k*(k+1)}}={\frac {k+2}{2(k+1)}}\\&(k+1)(k^{2}+2k)=k(k+1)(k+2)\\&k(k+1)(k+2)=k(k+1)(k+2)\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
