1 ∗ 3 1 ∗ 2 + 3 ∗ 3 2 2 ∗ 3 + ⋯ + ( 2 n − 1 ) 3 n n ( n + 1 ) = 3 n + 1 ( n + 1 ) − 3 {\displaystyle {\frac {1*3}{1*2}}+{\frac {3*3^{2}}{2*3}}+\cdots +{\frac {(2n-1)3^{n}}{n(n+1)}}={\frac {3^{n+1}}{(n+1)-3}}}
L : ( 2 n − 1 ) 3 n n ( n + 1 ) = 1 ∗ 3 2 = 1.5 R : 3 n + 1 ( n + 1 ) − 3 = 3 2 2 − 3 = 1.5 1 f r a c 1 2 = 1 1 2 √ {\displaystyle {\begin{aligned}&L:{\frac {(2n-1)3^{n}}{n(n+1)}}={\frac {1*3}{2}}=1.5\\&R:{\frac {3^{n+1}}{(n+1)-3}}={\frac {3^{2}}{2-3}}=1.5\\&1frac{1}{2}=1{\frac {1}{2}}\surd \\\end{aligned}}}
1 ∗ 3 1 ∗ 2 + 3 ∗ 3 2 2 ∗ 3 + ⋯ + ( 2 k − 1 ) 3 k k ( k + 1 ) = 3 k + 1 ( k + 1 ) − 3 {\displaystyle {\frac {1*3}{1*2}}+{\frac {3*3^{2}}{2*3}}+\cdots +{\frac {(2k-1)3^{k}}{k(k+1)}}={\frac {3^{k+1}}{(k+1)-3}}}
1 ∗ 3 1 ∗ 2 + 3 ∗ 3 2 2 ∗ 3 + ⋯ + ( 2 k − 1 ) 3 k k ( k + 1 ) ⏟ = 3 k + 1 ( k + 1 ) − 3 + ( 2 k + 1 ) 3 k + 1 ( k + 1 ) ( k + 2 ) = 3 k + 2 k + 2 − 3 3 k + 1 ( k + 1 ) − 3 + ( 2 k + 1 ) 3 k + 1 ( k + 1 ) ( k + 2 ) = 3 k + 2 k + 2 − 3 ( k + 2 ) ∗ 3 k + 1 + ( 2 k + 1 ) 3 k + 1 = ( k + 1 ) 3 k + 2 k + 2 + 2 k + 1 = 3 k + 3 3 k + 3 = 3 k + 3 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {1*3}{1*2}}+{\frac {3*3^{2}}{2*3}}+\cdots +{\frac {(2k-1)3^{k}}{k(k+1)}}} _{={\frac {3^{k+1}}{(k+1)-3}}}+{\frac {(2k+1)3^{k+1}}{(k+1)(k+2)}}={\frac {3^{k+2}}{k+2-3}}\\&{\frac {3^{k+1}}{(k+1)-3}}+{\frac {(2k+1)3^{k+1}}{(k+1)(k+2)}}={\frac {3^{k+2}}{k+2-3}}\\&(k+2)*3^{k+1}+(2k+1)3^{k+1}=(k+1)3^{k+2}\\&k+2+2k+1=3k+3\\&3k+3=3k+3\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
