1 2 + 2 4 + 3 9 + 4 16 + ⋯ + 2 n 2 2 n = 2 − n + 1 2 2 n − 1 {\displaystyle {\frac {1}{2}}+{\frac {2}{4}}+{\frac {3}{9}}+{\frac {4}{16}}+\cdots +{\frac {2n}{2^{2n}}}=2-{\frac {n+1}{2^{2n-1}}}}
L : − 2 n 2 2 n = 2 ∗ 1 2 2 ∗ 1 = 2 4 → 1 2 + 2 4 = 1 R : 2 − n + 1 2 2 n − 1 = 2 − 1 + 1 2 2 − 1 = 1 1 = 1 {\displaystyle {\begin{aligned}&L:-{\frac {2n}{2^{2n}}}={\frac {2*1}{2^{2*1}}}={\frac {2}{4}}\rightarrow {\frac {1}{2}}+{\frac {2}{4}}=1\\&R:2-{\frac {n+1}{2^{2n-1}}}=2-{\frac {1+1}{2^{2}-1}}=1\\&1=1\\\end{aligned}}}
1 2 + 2 4 + 3 9 + 4 16 + ⋯ + 2 k 2 2 k = 2 − k + 1 2 2 k − 1 {\displaystyle {\frac {1}{2}}+{\frac {2}{4}}+{\frac {3}{9}}+{\frac {4}{16}}+\cdots +{\frac {2k}{2^{2k}}}=2-{\frac {k+1}{2^{2k-1}}}}
1 2 + 2 4 + 3 9 + 4 16 + ⋯ + 2 k 2 2 k ⏟ = 2 − k + 1 2 2 k − 1 + 2 k + 1 2 2 k + 1 + 2 k + 2 2 2 k + 2 = 2 − k + 2 2 2 k + 1 2 − k + 1 2 2 k − 1 + 2 k + 1 2 2 k + 1 + 2 k + 2 2 2 k + 2 = 2 − k + 2 2 2 k + 1 2 − k + 1 2 2 k ∗ 2 − 1 + 2 k + 1 2 2 k ∗ 2 + 2 k + 2 2 2 k ∗ 2 2 = 2 − k + 2 2 2 k ∗ 2 / ∗ 2 2 k + 2 − 2 3 ( k + 1 ) + 2 ( 2 k + 1 ) + ( 2 k + 2 ) = − 2 ( 2 + k ) − 8 k − 8 + 4 k + 2 + 2 k + 2 = − 4 − 2 k − 2 k − 4 = − 4 − 2 k 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {1}{2}}+{\frac {2}{4}}+{\frac {3}{9}}+{\frac {4}{16}}+\cdots +{\frac {2k}{2^{2k}}}} _{=2-{\frac {k+1}{2^{2k-1}}}}+{\frac {2k+1}{2^{2k+1}}}+{\frac {2k+2}{2^{2k+2}}}=2-{\frac {k+2}{2^{2k+1}}}\\&2-{\frac {k+1}{2^{2k-1}}}+{\frac {2k+1}{2^{2k+1}}}+{\frac {2k+2}{2^{2k+2}}}=2-{\frac {k+2}{2^{2k+1}}}\\&2-{\frac {k+1}{2^{2k}*2^{-1}}}+{\frac {2k+1}{2^{2k}*2}}+{\frac {2k+2}{2^{2k}*2^{2}}}=2-{\frac {k+2}{2^{2k}*2}}/*2^{2k+2}\\&-2^{3}(k+1)+2(2k+1)+(2k+2)=-2(2+k)\\&-8k-8+4k+2+2k+2=-4-2k\\&-2k-4=-4-2k\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
