1 ∗ 1 + 2 ∗ 3 + 3 ∗ 9 + ⋯ + n ∗ 3 n − 1 = 1 4 [ 3 n ( 2 n − 1 ) + 1 ] {\displaystyle 1*1+2*3+3*9+\cdots +n*3^{n-1}={\frac {1}{4}}[3^{n}(2n-1)+1]}
L : n ∗ 3 n − 1 = 1 ∗ 3 0 = 1 R : 1 4 [ 3 ( 2 − 1 ) + 1 ] = 4 4 = 1 1 = 1 {\displaystyle {\begin{aligned}&L:n*3^{n-1}=1*3^{0}=1\\&R:{\frac {1}{4}}[3(2-1)+1]={\frac {4}{4}}=1\\&1=1\\\end{aligned}}}
1 ∗ 1 + 2 ∗ 3 + 3 ∗ 9 + ⋯ + k ∗ 3 k − 1 = 1 4 [ 3 k ( 2 k − 1 ) + 1 ] {\displaystyle 1*1+2*3+3*9+\cdots +k*3^{k-1}={\frac {1}{4}}[3^{k}(2k-1)+1]}
∗ 1 + 2 ∗ 3 + 3 ∗ 9 + ⋯ + k ∗ 3 k − 1 ⏟ = 1 4 [ 3 k ( 2 k − 1 ) + 1 ] + ( k + 1 ) ∗ 3 k = 1 4 [ 3 k + 1 ( 2 k + 1 ) + 1 ] 1 4 [ 3 k ( 2 k − 1 ) + 1 ] + ( k + 1 ) ∗ 3 k = 1 4 [ 3 k + 1 ( 2 k + 1 ) + 1 ] 1 4 [ 3 k ∗ 2 k − 3 k + 1 ] + 3 k ∗ k + 3 k = 1 4 [ 3 k + 1 ∗ 2 k + 3 k + 1 + 1 ] 3 k ∗ 2 k − 3 k + 1 + 4 k ∗ 3 k + 4 ∗ 3 k = 3 k + 1 ∗ 2 k + 3 k + 1 + 1 2 k − 1 + 4 k + 4 = 3 ∗ 2 k + 3 6 k + 3 = 6 k + 3 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {*1+2*3+3*9+\cdots +k*3^{k-1}} _{={\frac {1}{4}}[3^{k}(2k-1)+1]}+(k+1)*3^{k}={\frac {1}{4}}[3^{k+1}(2k+1)+1]\\&{\frac {1}{4}}[3^{k}(2k-1)+1]+(k+1)*3^{k}={\frac {1}{4}}[3^{k+1}(2k+1)+1]\\&{\frac {1}{4}}[3^{k}*2k-3^{k}+1]+3^{k}*k+3^{k}={\frac {1}{4}}[3^{k+1}*2k+3^{k+1}+1]\\&3^{k}*2k-3^{k}+1+4k*3^{k}+4*3^{k}=3^{k+1}*2^{k}+3^{k+1}+1\\&2k-1+4k+4=3*2^{k}+3\\&6k+3=6k+3\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
![{\displaystyle 1*1+2*3+3*9+\cdots +n*3^{n-1}={\frac {1}{4}}[3^{n}(2n-1)+1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e0f8cb158ed15258aafa8443f6fd969d7e5b424)
![{\displaystyle {\begin{aligned}&L:n*3^{n-1}=1*3^{0}=1\\&R:{\frac {1}{4}}[3(2-1)+1]={\frac {4}{4}}=1\\&1=1\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e105824c26870e4805beddd6893b1e99ccc2cb57)
טבעי![{\displaystyle 1*1+2*3+3*9+\cdots +k*3^{k-1}={\frac {1}{4}}[3^{k}(2k-1)+1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/581ed43ea6853fc8baed2f83f927742cc8557dff)
![{\displaystyle {\begin{aligned}&\underbrace {*1+2*3+3*9+\cdots +k*3^{k-1}} _{={\frac {1}{4}}[3^{k}(2k-1)+1]}+(k+1)*3^{k}={\frac {1}{4}}[3^{k+1}(2k+1)+1]\\&{\frac {1}{4}}[3^{k}(2k-1)+1]+(k+1)*3^{k}={\frac {1}{4}}[3^{k+1}(2k+1)+1]\\&{\frac {1}{4}}[3^{k}*2k-3^{k}+1]+3^{k}*k+3^{k}={\frac {1}{4}}[3^{k+1}*2k+3^{k+1}+1]\\&3^{k}*2k-3^{k}+1+4k*3^{k}+4*3^{k}=3^{k+1}*2^{k}+3^{k+1}+1\\&2k-1+4k+4=3*2^{k}+3\\&6k+3=6k+3\\&0=0\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/188f98064d295b25808245684cb67e96e4ff643f)
