( 2 n + 1 ) + ( 2 n + 2 ) + ( 2 n + 3 ) + ⋯ + ( 4 n ) = n ( 6 n + 1 ) {\displaystyle (2n+1)+(2n+2)+(2n+3)+\cdots +(4n)=n(6n+1)}
L : ( 4 n ) = 4 → ( 2 + 2 ) ( 2 + 2 ) = 7 R : n ( 6 n + 1 ) = ( 6 + 1 ) = 7 7 = 7 {\displaystyle {\begin{aligned}&L:(4n)=4\rightarrow (2+2)(2+2)=7\\&R:n(6n+1)=(6+1)=7\\&7=7\\\end{aligned}}}
( 2 k + 1 ) + ( 2 k + 2 ) + ( 2 k + 3 ) + ⋯ + ( 4 k ) = k ( 6 k + 1 ) {\displaystyle (2k+1)+(2k+2)+(2k+3)+\cdots +(4k)=k(6k+1)}
( 2 k + 3 ) + ( 2 k + 4 ) + ( 2 k + 5 ) + ⋯ + ( 4 k ) ⏟ = k ( 6 k + 1 ) − [ ( 2 k + 1 ) ( 2 k + 2 ) ] + ( 4 k + 1 ) + ( 4 k + 2 ) + ( 4 k + 3 ) + ( 4 k + 4 ) = ( k + 1 ) ( 6 k + 7 ) k ( 6 k + 1 ) − [ ( 2 k + 1 ) + ( 2 k + 2 ) ] + ( 4 k + 1 ) + ( 4 k + 2 ) + ( 4 k + 3 ) + ( 4 k + 4 ) = ( k + 1 ) ( 6 k + 7 ) 6 k 2 + k − 4 k − 3 + 4 k + 1 + 4 k + 2 + 4 k + 3 + 4 k + 4 = 6 k 2 + 7 k + 6 k + 7 6 k 2 + 13 k + 7 = 6 k 2 + 13 k + 7 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {(2k+3)+(2k+4)+(2k+5)+\cdots +(4k)} _{=k(6k+1)-[(2k+1)(2k+2)]}+(4k+1)+(4k+2)+(4k+3)+(4k+4)=(k+1)(6k+7)\\&k(6k+1)-[(2k+1)+(2k+2)]+(4k+1)+(4k+2)+(4k+3)+(4k+4)=(k+1)(6k+7)\\&6k^{2}+k-4k-3+4k+1+4k+2+4k+3+4k+4=6k^{2}+7k+6k+7\\&6k^{2}+13k+7=6k^{2}+13k+7\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
![{\displaystyle {\begin{aligned}&\underbrace {(2k+3)+(2k+4)+(2k+5)+\cdots +(4k)} _{=k(6k+1)-[(2k+1)(2k+2)]}+(4k+1)+(4k+2)+(4k+3)+(4k+4)=(k+1)(6k+7)\\&k(6k+1)-[(2k+1)+(2k+2)]+(4k+1)+(4k+2)+(4k+3)+(4k+4)=(k+1)(6k+7)\\&6k^{2}+k-4k-3+4k+1+4k+2+4k+3+4k+4=6k^{2}+7k+6k+7\\&6k^{2}+13k+7=6k^{2}+13k+7\\&0=0\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9059ca443585795da524bcb7e32a3c253925db2a)
