15 + 48 + 105 + ⋯ + n ( n + 2 ) ( n + 4 ) = n 4 ( n + 1 ) ( n + 4 ) ( n + 5 ) {\displaystyle 15+48+105+\cdots +n(n+2)(n+4)={\frac {n}{4}}(n+1)(n+4)(n+5)}
L : n ( n + 2 ) ( n + 4 ) → 1 ( 1 + 2 ) ( 1 + 4 ) = 15 R : n 4 ( n + 1 ) ( n + 4 ) ( n + 5 ) → 1 4 ( 1 + 1 ) ( 1 + 4 ) ( 1 + 5 ) = 15 15 = 15 √ {\displaystyle {\begin{aligned}&L:n(n+2)(n+4)\rightarrow 1(1+2)(1+4)=15\\&R:{\frac {n}{4}}(n+1)(n+4)(n+5)\rightarrow {\frac {1}{4}}(1+1)(1+4)(1+5)=15\\&15=15\surd \\\end{aligned}}}
15 + 48 + 105 + ⋯ + k ( k + 2 ) ( k + 4 ) = k 4 ( k + 1 ) ( k + 4 ) ( k + 5 ) {\displaystyle 15+48+105+\cdots +k(k+2)(k+4)={\frac {k}{4}}(k+1)(k+4)(k+5)}
15 + 48 + 105 + ⋯ + k ( k + 2 ) ( k + 4 ) ⏟ = k 4 ( k + 1 ) ( k + 4 ) ( k + 5 ) + ( k + 1 ) ( k + 3 ) ( k + 5 ) = k + 1 4 ( k + 2 ) ( k + 5 ) ( k + 6 ) / ∗ 4 ( k + 1 ) ( k + 5 ) k ( k + 4 ) + 4 ( k + 3 ) = ( k + 2 ) ( k + 6 ) k 2 + 4 k + 4 k + 12 = k 2 + 8 k + 12 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {15+48+105+\cdots +k(k+2)(k+4)} _{={\frac {k}{4}}(k+1)(k+4)(k+5)}+{\color {red}(k+1)(k+3)(k+5)}={\frac {k+1}{4}}(k+2)(k+5)(k+6)/*{\frac {4}{(k+1)(k+5)}}\\&k(k+4)+4(k+3)=(k+2)(k+6)\\&k^{2}+4k+4k+12=k^{2}+8k+12\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.