1 ∗ 2 + 3 ∗ 4 + 5 ∗ 6 + ⋯ + ( 2 n − 1 ) 2 n = n 3 ( n + 1 ) ( 4 n − 1 ) {\displaystyle 1*2+3*4+5*6+\cdots +(2n-1)2n={\frac {n}{3}}(n+1)(4n-1)}
L : ( 2 n − 1 ) 2 n ⇒ ( 2 ∗ 1 − 1 ) 2 ∗ 1 = 2 ⏟ 1 ∗ 2 = 2 R : n 3 ( n + 1 ) ( 4 n − 1 ) ⇒ 1 3 ( 1 + 1 ) ( 4 ∗ 1 − 1 ) = 2 2 = 2 √ {\displaystyle {\begin{aligned}L:(2n-1)2n\Rightarrow (2*1-1)2*1=\underbrace {2} _{1*2=2}\\R:{\frac {n}{3}}(n+1)(4n-1)\Rightarrow {\frac {1}{3}}(1+1)(4*1-1)=2\\2=2\surd \\\end{aligned}}}
1 ∗ 2 + 3 ∗ 4 + 5 ∗ 6 + ⋯ + ( 2 k − 1 ) 2 k = k 3 ( k + 1 ) ( 4 k − 1 ) {\displaystyle 1*2+3*4+5*6+\cdots +(2k-1)2k={\frac {k}{3}}(k+1)(4k-1)}
1 ∗ 2 + 3 ∗ 4 + 5 ∗ 6 + ⋯ + ( 2 k − 1 ) 2 k ⏟ = k 3 ( k + 1 ) ( 4 k − 1 ) + [ 2 ( k + 1 ) − 1 ] ∗ 2 ( k + 1 ) = k + 1 3 ( k + 1 + 1 ) [ 4 ( k + 1 ) − 1 ] k 3 ( k + 1 ) ( 4 k − 1 ) + ( 2 k + 2 − 1 ) ∗ 2 ( k + 1 ) = k + 1 3 ( k + 1 + 1 ) ( 4 k + 4 − 1 ) k 3 ( k + 1 ) ( 4 k − 1 ) + ( 2 k + 1 ) ∗ 2 ( k + 1 ) = k + 1 3 ( k + 2 ) ( 4 k + 3 ) / : ( k + 1 ) k 3 ( 4 k − 1 ) + ( 2 k + 1 ) ∗ 2 = 1 3 ( k + 2 ) ( 4 k + 3 ) / ∗ 3 k ( 4 k − 1 ) + 6 ( 2 k + 1 ) = ( k + 2 ) ( 4 k + 3 ) 4 k 2 − k + 12 k + 6 = 4 k 2 + 3 k + 8 k + 6 0 = 0 √ {\displaystyle {\begin{aligned}\underbrace {1*2+3*4+5*6+\cdots +(2k-1)2k} _{={\frac {k}{3}}(k+1)(4k-1)}+{\color {red}[2(k+1)-1]*2(k+1)}={\color {red}{\frac {k+1}{3}}(k+1+1)[4(k+1)-1]}\\{\frac {k}{3}}(k+1)(4k-1)+(2k+2-1)*2(k+1)={\frac {k+1}{3}}(k+1+1)(4k+4-1)\\{\frac {k}{3}}(k+1)(4k-1)+(2k+1)*2(k+1)={\frac {k+1}{3}}(k+2)(4k+3)/:(k+1)\\{\frac {k}{3}}(4k-1)+(2k+1)*2={\frac {1}{3}}(k+2)(4k+3)/*3\\k(4k-1)+6(2k+1)=(k+2)(4k+3)\\4k^{2}-k+12k+6=4k^{2}+3k+8k+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*4+5*6+\cdots +(2k-1)2k} _{={\frac {k}{3}}(k+1)(4k-1)}+{\color {red}[2(k+1)-1]*2(k+1)}={\color {red}{\frac {k+1}{3}}(k+1+1)[4(k+1)-1]}\\{\frac {k}{3}}(k+1)(4k-1)+(2k+2-1)*2(k+1)={\frac {k+1}{3}}(k+1+1)(4k+4-1)\\{\frac {k}{3}}(k+1)(4k-1)+(2k+1)*2(k+1)={\frac {k+1}{3}}(k+2)(4k+3)/:(k+1)\\{\frac {k}{3}}(4k-1)+(2k+1)*2={\frac {1}{3}}(k+2)(4k+3)/*3\\k(4k-1)+6(2k+1)=(k+2)(4k+3)\\4k^{2}-k+12k+6=4k^{2}+3k+8k+6\\0=0\surd \\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3a167ad785878169e147def4c45aff110202666)
