2 + 4 + 6 + ⋯ + 2 n = n ( n + 1 ) {\displaystyle 2+4+6+\cdots +2n=n(n+1)}
2 ∗ 1 = 1 ( 1 + 1 ) 2 = 2 √ {\displaystyle {\begin{aligned}2*1=1(1+1)\\2=2\surd \\\end{aligned}}}
2 + 4 + 6 + ⋯ + 2 k = k ( k + 1 ) {\displaystyle 2+4+6+\cdots +2k=k(k+1)}
2 + 4 + 6 + ⋯ + 2 k ⏟ = k ( k + 1 ) + 2 ( k + 1 ) = ( k + 1 ) ( k + 2 ) k ( k + 1 ) + 2 ( k + 1 ) = ( k + 1 ) ( k + 2 ) / : ( k + 1 ) k + 2 = k + 2 {\displaystyle {\begin{aligned}\underbrace {2+4+6+\cdots +2k} _{=k(k+1)}+{\color {red}2(k+1)}={\color {red}(k+1)(k+2)}\\k(k+1)+2(k+1)=(k+1)(k+2)/:(k+1)\\k+2=k+2\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.