1 + 2 2 + 3 4 + ⋯ + n 2 n − 1 = 4 − n + 2 2 n − 1 {\displaystyle 1+{\frac {2}{2}}+{\frac {3}{4}}+\cdots +{\frac {n}{2^{n-1}}}=4-{\frac {n+2}{2^{n-1}}}}
L : n 2 n − 1 = 1 2 1 − 1 = 1 R : 4 − n + 2 2 n − 1 = 4 − 1 + 2 2 1 − 1 = 4 − 3 = 1 1 = 1 √ {\displaystyle {\begin{aligned}&L:{\frac {n}{2^{n-1}}}={\frac {1}{2^{1-1}}}=1\\&R:4-{\frac {n+2}{2^{n-1}}}=4-{\frac {1+2}{2^{1-1}}}=4-3=1\\&1=1\surd \\\end{aligned}}}
1 + 2 2 + 3 4 + ⋯ + k 2 k − 1 = 4 − k + 2 2 k − 1 {\displaystyle 1+{\frac {2}{2}}+{\frac {3}{4}}+\cdots +{\frac {k}{2^{k-1}}}=4-{\frac {k+2}{2^{k-1}}}}
1 + 2 2 + 3 4 + ⋯ + k 2 k − 1 ⏟ = 4 − k + 2 2 k − 1 + k + 1 2 k = 4 − k + 3 2 k 4 − k + 2 2 k − 1 + k + 1 2 k = 4 − k + 3 2 k / ∗ 2 k − 4 − 2 ( k + 2 ) + 9 k + 1 ) = − ( k + 3 ) − 2 k − 4 + k + 1 = − k + 3 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {1+{\frac {2}{2}}+{\frac {3}{4}}+\cdots +{\frac {k}{2^{k-1}}}} _{=4-{\frac {k+2}{2^{k-1}}}}+{\frac {k+1}{2^{k}}}=4-{\frac {k+3}{2^{k}}}\\&4-{\frac {k+2}{2^{k-1}}}+{\frac {k+1}{2^{k}}}=4-{\frac {k+3}{2^{k}}}/*{\frac {2^{k}}{-4}}\\&-2(k+2)+9k+1)=-(k+3)\\&-2k-4+k+1=-k+3\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.