1 3 + 1 9 + 1 27 + ⋯ + 1 3 n = 1 2 − 1 2 ∗ 3 n {\displaystyle {\frac {1}{3}}+{\frac {1}{9}}+{\frac {1}{27}}+\cdots +{\frac {1}{3^{n}}}={\frac {1}{2}}-{\frac {1}{2*3^{n}}}}
L : 1 3 n = 1 3 1 = 1 3 R : 1 2 − 1 2 ∗ 3 n = 1 2 − 1 6 = 1 3 1 3 = 1 3 √ {\displaystyle {\begin{aligned}&L:{\frac {1}{3^{n}}}={\frac {1}{3^{1}}}={\frac {1}{3}}\\&R:{\frac {1}{2}}-{\frac {1}{2*3^{n}}}={\frac {1}{2}}-{\frac {1}{6}}={\frac {1}{3}}\\&{\frac {1}{3}}={\frac {1}{3}}\surd \\\end{aligned}}}
1 3 + 1 9 + 1 27 + ⋯ + 1 3 k = 1 2 − 1 2 ∗ 3 k {\displaystyle {\frac {1}{3}}+{\frac {1}{9}}+{\frac {1}{27}}+\cdots +{\frac {1}{3^{k}}}={\frac {1}{2}}-{\frac {1}{2*3^{k}}}}
1 3 + 1 9 + 1 27 + ⋯ + 1 3 k ⏟ = 1 2 − 1 2 ∗ 3 k + 1 3 k + 1 = 1 2 − 1 2 ∗ 3 k + 1 1 2 − 1 2 ∗ 3 k + 1 3 k + 1 = 1 2 − 1 2 ∗ 3 k + 1 / ∗ 2 ∗ 3 k + 1 3 k + 1 − 3 + 2 = 3 k + 1 − 1 3 k + 1 − 1 = 3 k + 1 − 1 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {1}{3}}+{\frac {1}{9}}+{\frac {1}{27}}+\cdots +{\frac {1}{3^{k}}}} _{={\frac {1}{2}}-{\frac {1}{2*3^{k}}}}+{\frac {1}{3^{k+1}}}={\frac {1}{2}}-{\frac {1}{2*3^{k+1}}}\\&{\frac {1}{2}}-{\frac {1}{2*3^{k}}}+{\frac {1}{3^{k+1}}}={\frac {1}{2}}-{\frac {1}{2*3^{k+1}}}/*2*3^{k+1}\\&3^{k+1}-3+2=3^{k+1}-1\\&3^{k+1}-1=3^{k+1}-1\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
