3 ∗ 2 + 6 ∗ 3 + 9 ∗ 4 + ⋯ + 3 n ( n + 1 ) = n ( n + 1 ) ( n + 2 ) {\displaystyle 3*2+6*3+9*4+\cdots +3n(n+1)=n(n+1)(n+2)}
( 1 + 1 ) = 1 ( 1 + 1 ) ( 1 + 2 ) 2 ⏟ 3 ∗ 2 = 2 ∗ 3 √ {\displaystyle {\begin{aligned}(1+1)=1(1+1)(1+2)\\\underbrace {2} _{3*2}=2*3\surd \\\end{aligned}}}
3 ∗ 2 + 6 ∗ 3 + 9 ∗ 4 + ⋯ + 3 k ( k + 1 ) = k ( k + 1 ) ( k + 2 ) {\displaystyle 3*2+6*3+9*4+\cdots +3k(k+1)=k(k+1)(k+2)}
3 ∗ 2 + 6 ∗ 3 + 9 ∗ 4 + ⋯ + 3 k ( k + 1 ) ⏟ = k ( k + 1 ) ( k + 2 ) + 3 ( k + 1 ) ( k + 1 + 1 ) = ( k + 1 ) ( k + 2 ) ( k + 3 ) k ( k + 1 ) ( k + 2 ) + 3 ( k + 1 ) ( k + 2 ) = ( k + 1 ) ( k + 2 ) ( k + 3 ) / : ( k + 1 ) ( k + 2 ) k + 3 = k + 3 0 = 0 √ {\displaystyle {\begin{aligned}\underbrace {3*2+6*3+9*4+\cdots +3k(k+1)} _{=k(k+1)(k+2)}+{\color {red}3(k+1)(k+1+1)}={\color {red}(k+1)(k+2)(k+3)}\\k(k+1)(k+2)+3(k+1)(k+2)=(k+1)(k+2)(k+3)/:(k+1)(k+2)\\k+3=k+3\\0=0\surd \\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
