( 1 + 3 1 ) ( 1 + 5 4 ) ( 1 + 7 9 ) + ⋯ + ( 1 + 2 n + 1 n 2 = ( n + 1 ) 2 {\displaystyle (1+{\frac {3}{1}})(1+{\frac {5}{4}})(1+{\frac {7}{9}})+\cdots +(1+{\frac {2n+1}{n^{2}}}=(n+1)^{2}}
L : ( 1 + 2 n + 1 n 2 = 1 + 2 ∗ 1 + 1 1 2 = 1 + 3 1 = 4 R : ( n + 1 ) 2 = ( 1 + 1 ) 2 = 4 4 = 4 {\displaystyle {\begin{aligned}&L:(1+{\frac {2n+1}{n^{2}}}=1+{\frac {2*1+1}{1^{2}}}=1+{\frac {3}{1}}=4\\&R:(n+1)^{2}=(1+1)^{2}=4\\&4=4\\\end{aligned}}}
( 1 + 3 1 ) ( 1 + 5 4 ) ( 1 + 7 9 ) + ⋯ + ( 1 + 2 k + 1 k 2 = ( k + 1 ) 2 {\displaystyle (1+{\frac {3}{1}})(1+{\frac {5}{4}})(1+{\frac {7}{9}})+\cdots +(1+{\frac {2k+1}{k^{2}}}=(k+1)^{2}}
( 1 + 3 1 ) ( 1 + 5 4 ) ( 1 + 7 9 ) + ⋯ + ( 1 + 2 k + 1 k 2 ) ⏟ = ( k + 1 ) 2 ∗ ( 1 + 2 k + 3 ( k + 1 ) 2 = ( k + 2 ) 2 ( k + 1 ) 2 ∗ ( 1 + 2 k + 3 ( k + 1 ) 2 ) = ( k + 2 ) 2 ( k + 1 ) 2 ∗ ( k + 1 ) 2 + 2 k + 3 ( k + 1 ) 2 = ( k + 2 ) 2 k 2 + 2 k + 1 + 2 k + 3 = k 2 + 4 k + 4 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {(1+{\frac {3}{1}})(1+{\frac {5}{4}})(1+{\frac {7}{9}})+\cdots +(1+{\frac {2k+1}{k^{2}}})} _{=(k+1)^{2}}*(1+{\frac {2k+3}{(k+1)^{2}}}=(k+2)^{2}\\&(k+1)^{2}*(1+{\frac {2k+3}{(k+1)^{2}}})=(k+2)^{2}\\&(k+1)^{2}*{\frac {(k+1)^{2}+2k+3}{(k+1)^{2}}}=(k+2)^{2}\\&k^{2}+2k+1+2k+3=k^{2}+4k+4\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
