4 + 5 + 6 + ⋯ + ( n + 3 ) = n 2 ( n + 7 ) {\displaystyle 4+5+6+\cdots +(n+3)={\frac {n}{2}}(n+7)}
( 1 + 3 ) = 1 2 ( 1 + 7 ) 4 = 4 √ {\displaystyle {\begin{aligned}(1+3)={\frac {1}{2}}(1+7)\\4=4\surd \\\end{aligned}}}
4 + 5 + 6 + ⋯ + ( k + 3 ) = k 2 ( k + 7 ) {\displaystyle 4+5+6+\cdots +(k+3)={\frac {k}{2}}(k+7)}
4 + 5 + 6 + ⋯ + ( k + 3 ) ⏟ = k 2 ( k + 7 ) + ( k + 1 + 3 ) = k + 1 2 ( k + 1 + 7 ) k 2 ( k + 7 ) + ( k + 4 ) = k + 1 2 ( k + 8 ) / ∗ 2 k ( k + 7 ) + 2 ( k + 4 ) = ( k + 1 ) ( k + 8 ) k 2 + 7 k + 2 k + 8 = k 2 + 9 k + 8 0 = 0 √ {\displaystyle {\begin{aligned}\underbrace {4+5+6+\cdots +(k+3)} _{={\frac {k}{2}}(k+7)}+{\color {red}(k+1+3)}={\color {red}{\frac {k+1}{2}}(k+1+7)}\\{\frac {k}{2}}(k+7)+(k+4)={\frac {k+1}{2}}(k+8)/*2\\k(k+7)+2(k+4)=(k+1)(k+8)\\k^{2}+7k+2k+8=k^{2}+9k+8\\0=0\surd \\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.