6 + 16 + 42 + ⋯ + ( 4 ∗ 3 n − 1 + 2 n ) = 2 ∗ 3 n + ( n − 1 ) ( n + 2 ) {\displaystyle 6+16+42+\cdots +(4*3^{n-1}+2n)=2*3^{n}+(n-1)(n+2)}
L : 4 ∗ 3 0 + 2 ∗ 1 = 6 R : 2 ∗ 3 ( 1 − 1 ) ( 1 + 2 ) = 6 6 = 6 √ {\displaystyle {\begin{aligned}&L:4*3^{0}+2*1=6\\&R:2*3(1-1)(1+2)=6\\&6=6\surd \\\end{aligned}}}
6 + 16 + 42 + ⋯ + ( 4 ∗ 3 k − 1 + 2 k ) = 2 ∗ 3 k + ( k − 1 ) ( k + 2 ) {\displaystyle 6+16+42+\cdots +(4*3^{k-1}+2k)=2*3^{k}+(k-1)(k+2)}
6 + 16 + 42 + ⋯ + ( 4 ∗ 3 k − 1 + 2 k ) ⏟ = 2 ∗ 3 k + ( k − 1 ) ( k + 2 ) + ( 4 ∗ 3 k + 2 k + 2 ) = 2 ∗ 3 k + 1 + ( k ) ( k + 3 ) 2 ∗ 3 k + ( k − 1 ) ( k + 2 ) + ( 4 ∗ 3 k + 2 k + 2 ) = 2 ∗ 3 k + 1 + ( k ) ( k + 3 ) 2 ∗ 3 k + k 2 + k − 2 + 4 ∗ 3 k + 2 k + 2 = 2 ∗ 3 k + 1 + k 2 + 3 k 2 ∗ 3 k + 4 ∗ 3 k = 2 ∗ 3 k + 1 / : 3 k 2 + 4 = 2 ∗ 3 6 = 6 {\displaystyle {\begin{aligned}&\underbrace {6+16+42+\cdots +(4*3^{k-1}+2k)} _{=2*3^{k}+(k-1)(k+2)}+(4*3^{k}+2k+2)=2*3^{k+1}+(k)(k+3)\\&2*3^{k}+(k-1)(k+2)+(4*3^{k}+2k+2)=2*3^{k+1}+(k)(k+3)\\&2*3^{k}+k^{2}+k-2+4*3^{k}+2k+2=2*3^{k+1}+k^{2}+3k\\&2*3^{k}+4*3^{k}=2*3^{k+1}/:3^{k}\\&2+4=2*3\\&6=6\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
