1 ∗ 1 + 2 ∗ 3 + 3 ∗ 9 + 4 ∗ 27 + ⋯ + 2 n ∗ 3 2 n − 1 = 1 4 [ ( 4 n − 1 ) ∗ 3 2 n + 1 ] {\displaystyle 1*1+2*3+3*9+4*27+\cdots +2n*3^{2n-1}={\frac {1}{4}}[(4n-1)*3^{2n}+1]}
L : 2 n ∗ 3 2 n − 1 = 3 ∗ 1 ∗ 3 2 − 1 = 2 ∗ 3 → 1 ∗ 1 + 2 ∗ 3 = 7 R : 1 4 [ ( 4 n − 1 ) ∗ 3 2 n + 1 ] = 1 4 [ ( 4 − 1 ) 3 2 + 1 ] = 7 7 = 7 {\displaystyle {\begin{aligned}&L:2n*3^{2n-1}=3*1*3^{2-1}=2*3\rightarrow 1*1+2*3=7\\&R:{\frac {1}{4}}[(4n-1)*3^{2n}+1]={\frac {1}{4}}[(4-1)3^{2}+1]=7\\&7=7\\\end{aligned}}}
1 ∗ 1 + 2 ∗ 3 + 3 ∗ 9 + 4 ∗ 27 + ⋯ + 2 k ∗ 3 2 k − 1 = 1 4 [ ( 4 k − 1 ) ∗ 3 2 k + 1 ] {\displaystyle 1*1+2*3+3*9+4*27+\cdots +2k*3^{2k-1}={\frac {1}{4}}[(4k-1)*3^{2k}+1]}
1 ∗ 1 + 2 ∗ 3 + 3 ∗ 9 + 4 ∗ 27 + ⋯ + 2 k ∗ 3 2 k − 1 ⏟ = 1 4 [ ( 4 k − 1 ) ∗ 3 2 k + 1 ] + ( 2 k + 1 ) ∗ 3 2 k + ( 2 k + 2 ) ∗ 3 2 k + 1 = 1 4 [ ( 4 k + 3 ) ∗ 3 2 k + 2 + 1 ] 1 4 [ ( 4 k − 1 ) ∗ 3 2 k + 1 ] + ( 2 k + 1 ) ∗ 3 2 k + ( 2 k + 2 ) ∗ 3 2 k + 1 = 1 4 [ ( 4 k + 3 ) ∗ 3 2 k + 2 + 1 ] ( 4 k − 1 ) ∗ 3 2 k + 1 + 4 ∗ ( 2 k + 1 ) ∗ 3 2 k + 4 ∗ ( 2 k + 2 ) ∗ 3 2 k + 1 = ( 4 k + 3 ) ∗ 3 2 k + 2 + 1 ( 4 k − 1 ) ∗ 3 2 k + 4 ∗ ( 2 k + 1 ) ∗ 3 2 k + 4 ∗ ( 2 k + 2 ) ∗ 3 2 k + 1 = ( 4 k + 3 ) ∗ 3 2 k + 2 / : 3 2 k 4 k − 1 + 4 ( 2 k + 1 ) + 4 ( 2 k + 2 ) ∗ 3 = ( 4 k + 3 ) ∗ 3 2 4 k − 1 + 8 k + 4 + 24 k + 24 = 36 k + 27 36 k + 27 = 36 k + 27 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {1*1+2*3+3*9+4*27+\cdots +2k*3^{2k-1}} _{={\frac {1}{4}}[(4k-1)*3^{2k}+1]}+(2k+1)*3^{2k}+(2k+2)*3^{2k+1}={\frac {1}{4}}[(4k+3)*3^{2k+2}+1]\\&{\frac {1}{4}}[(4k-1)*3^{2k}+1]+(2k+1)*3^{2k}+(2k+2)*3^{2k+1}={\frac {1}{4}}[(4k+3)*3^{2k+2}+1]\\&(4k-1)*3^{2k}+1+4*(2k+1)*3^{2k}+4*(2k+2)*3^{2k+1}=(4k+3)*3^{2k+2}+1\\&(4k-1)*3^{2k}+4*(2k+1)*3^{2k}+4*(2k+2)*3^{2k+1}=(4k+3)*3^{2k+2}/:3^{2k}\\&4k-1+4(2k+1)+4(2k+2)*3=(4k+3)*3^{2}\\&4k-1+8k+4+24k+24=36k+27\\&36k+27=36k+27\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
![{\displaystyle 1*1+2*3+3*9+4*27+\cdots +2n*3^{2n-1}={\frac {1}{4}}[(4n-1)*3^{2n}+1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c7dd9b798dde28c2a48093778c623006a319e06)
![{\displaystyle {\begin{aligned}&L:2n*3^{2n-1}=3*1*3^{2-1}=2*3\rightarrow 1*1+2*3=7\\&R:{\frac {1}{4}}[(4n-1)*3^{2n}+1]={\frac {1}{4}}[(4-1)3^{2}+1]=7\\&7=7\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b120bb6f1bd145c2299b562d9abf74f54347b9e3)
טבעי![{\displaystyle 1*1+2*3+3*9+4*27+\cdots +2k*3^{2k-1}={\frac {1}{4}}[(4k-1)*3^{2k}+1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fffba6ca732163300cc1b9fc85d0a88194f20543)
![{\displaystyle {\begin{aligned}&\underbrace {1*1+2*3+3*9+4*27+\cdots +2k*3^{2k-1}} _{={\frac {1}{4}}[(4k-1)*3^{2k}+1]}+(2k+1)*3^{2k}+(2k+2)*3^{2k+1}={\frac {1}{4}}[(4k+3)*3^{2k+2}+1]\\&{\frac {1}{4}}[(4k-1)*3^{2k}+1]+(2k+1)*3^{2k}+(2k+2)*3^{2k+1}={\frac {1}{4}}[(4k+3)*3^{2k+2}+1]\\&(4k-1)*3^{2k}+1+4*(2k+1)*3^{2k}+4*(2k+2)*3^{2k+1}=(4k+3)*3^{2k+2}+1\\&(4k-1)*3^{2k}+4*(2k+1)*3^{2k}+4*(2k+2)*3^{2k+1}=(4k+3)*3^{2k+2}/:3^{2k}\\&4k-1+4(2k+1)+4(2k+2)*3=(4k+3)*3^{2}\\&4k-1+8k+4+24k+24=36k+27\\&36k+27=36k+27\\&0=0\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d915e6fa00ba5da9b3994c4a581b7f0877d13f60)
