3∗2+6∗3+9∗4+⋯+3n(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+⋯+3k(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+⋯+3k(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+30=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.