1 ∗ 2 ∗ 3 + 2 ∗ 3 ∗ 5 + ⋯ + n ( n + 1 ) ( 2 n + 1 ) = n 2 ( n + 1 ) 2 ( n + 2 ) {\displaystyle 1*2*3+2*3*5+\cdots +n(n+1)(2n+1)={\frac {n}{2}}(n+1)^{2}(n+2)}
L : ( n + 1 ) ( 2 n + 1 ) ⇒ ( 1 + 1 ) ( 2 ∗ 1 + 1 ) = 3 ⏟ 1 ∗ 2 ∗ 3 = 6 R : n 2 ( n + 1 ) 2 ( n + 2 ) ⇒ 1 2 ( 1 + 1 ) 2 ( 1 + 2 ) = 6 6 = 6 √ {\displaystyle {\begin{aligned}L:(n+1)(2n+1)\Rightarrow (1+1)(2*1+1)=\underbrace {3} _{1*2*3=6}\\R:{\frac {n}{2}}(n+1)^{2}(n+2)\Rightarrow {\frac {1}{2}}(1+1)^{2}(1+2)=6\\6=6\surd \\\end{aligned}}}
1 ∗ 2 ∗ 3 + 2 ∗ 3 ∗ 5 + ⋯ + k ( k + 1 ) ( 2 k + 1 ) = k 2 ( k + 1 ) 2 ( k + 2 ) {\displaystyle 1*2*3+2*3*5+\cdots +k(k+1)(2k+1)={\frac {k}{2}}(k+1)^{2}(k+2)}
1 ∗ 2 ∗ 3 + 2 ∗ 3 ∗ 5 + ⋯ + k ( k + 1 ) ( 2 k + 1 ) ⏟ = k 2 ( k + 1 ) 2 ( k + 2 ) + ( k + 1 ) ( k + 2 ) [ 2 ( k + 1 ) + 1 ] = k + 1 2 ( k + 2 ) 2 ∗ ( k + 3 ) k 2 ( k + 1 ) 2 ( k + 2 ) + ( k + 1 ) ( k + 2 ) ( 2 k + 3 ) = k + 1 2 ( k + 2 ) 2 ∗ ( k + 3 ) / ∗ 2 ( k + 1 ) ( k + 2 ) k ( k + 1 ) + 2 ( 2 k + 3 ) = ( k + 2 ) ( k + 3 ) k 2 + k + 4 k + 6 = k 2 + 3 k + 2 k + 6 0 = 0 √ {\displaystyle {\begin{aligned}\underbrace {1*2*3+2*3*5+\cdots +k(k+1)(2k+1)} _{={\frac {k}{2}}(k+1)^{2}(k+2)}+{\color {red}(k+1)(k+2)[2(k+1)+1]}={\color {red}{\frac {k+1}{2}}(k+2)^{2}*(k+3)}\\{\frac {k}{2}}(k+1)^{2}(k+2)+(k+1)(k+2)(2k+3)={\frac {k+1}{2}}(k+2)^{2}*(k+3)/*{\frac {2}{(k+1)(k+2)}}\\k(k+1)+2(2k+3)=(k+2)(k+3)\\k^{2}+k+4k+6=k^{2}+3k+2k+6\\0=0\surd \\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
![{\displaystyle {\begin{aligned}\underbrace {1*2*3+2*3*5+\cdots +k(k+1)(2k+1)} _{={\frac {k}{2}}(k+1)^{2}(k+2)}+{\color {red}(k+1)(k+2)[2(k+1)+1]}={\color {red}{\frac {k+1}{2}}(k+2)^{2}*(k+3)}\\{\frac {k}{2}}(k+1)^{2}(k+2)+(k+1)(k+2)(2k+3)={\frac {k+1}{2}}(k+2)^{2}*(k+3)/*{\frac {2}{(k+1)(k+2)}}\\k(k+1)+2(2k+3)=(k+2)(k+3)\\k^{2}+k+4k+6=k^{2}+3k+2k+6\\0=0\surd \\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87d142d852f2aa247daa1326d5df6be4d3101c85)
