1∗2∗3∗4+2∗3∗4∗5+⋯+n(n+1)(n+2)(n+3)=n5(n+1)(n+2)(n+3)(n+4){\displaystyle 1*2*3*4+2*3*4*5+\cdots +n(n+1)(n+2)(n+3)={\frac {n}{5}}{\color {red}(n+1)}(n+2)(n+3)(n+4)}
R:n(n+1)(n+2)(n+3)=1∗2∗3∗4=24L:n5(n+1)(n+2)(n+3)(n+4)=1∗2∗3∗4∗55=2424=24√{\displaystyle {\begin{aligned}&R:n(n+1)(n+2)(n+3)=1*2*3*4=24\\&L:{\frac {n}{5}}(n+1)(n+2)(n+3)(n+4)={\frac {1*2*3*4*5}{5}}=24\\&24=24\surd \\\end{aligned}}}
1∗2∗3∗4+2∗3∗4∗5+⋯+k(k+1)(k+2)(k+3)=k5(k+1)(k+2)(k+3)(k+4){\displaystyle 1*2*3*4+2*3*4*5+\cdots +k(k+1)(k+2)(k+3)={\frac {k}{5}}(k+1)(k+2)(k+3)(k+4)}
1∗2∗3∗4+2∗3∗4∗5+⋯+k(k+1)(k+2)(k+3)⏟=k5(k+1)(k+2)(k+3)(k+4)+(k+1)(k+2)(k+3)(k+4)=(k+1)5(k+2)(k+3)(k+4)(k+5)k5(k+1)(k+2)(k+3)(k+4)+(k+1)(k+2)(k+3)(k+4)=(k+1)5(k+2)(k+3)(k+4)(k+5)/∗5(k+1)(k+2)(k+3)(k+4)k+5=k+50=0{\displaystyle {\begin{aligned}&\underbrace {1*2*3*4+2*3*4*5+\cdots +k(k+1)(k+2)(k+3)} _{={\frac {k}{5}}(k+1)(k+2)(k+3)(k+4)}+{\color {red}(k+1)(k+2)(k+3)(k+4)}={\frac {(k+1)}{5}}(k+2)(k+3)(k+4)(k+5)\\&{\frac {k}{5}}(k+1)(k+2)(k+3)(k+4)+(k+1)(k+2)(k+3)(k+4)={\frac {(k+1)}{5}}(k+2)(k+3)(k+4)(k+5)/*{\frac {5}{(k+1)(k+2)(k+3)(k+4)}}\\&k+5=k+5\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.