cho s = 1/3 – 2/ 3^2 + 3/ 3^3 +…..+99/3^99 -100/3^100 so sánh s và 1/5 26/08/2021 Bởi Autumn cho s = 1/3 – 2/ 3^2 + 3/ 3^3 +…..+99/3^99 -100/3^100 so sánh s và 1/5
$S=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}+\cdots+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\\ 3S=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+\cdots+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ S+3S=1-\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{3}{3^2}-\dfrac{2}{3^2}\right)-\left(\dfrac{4}{3^3}-\dfrac{3}{3^3}\right)+\cdots-\left(\dfrac{100}{3^{99}}-\dfrac{99}{3^{99}}\right)-\dfrac{100}{3^{100}}\\ \Leftrightarrow 4S=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\cdots-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\\ 12S=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-\cdots-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\\ 4S+12S=3-\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Leftrightarrow 16S=3-\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}<3\\ \Rightarrow S<\dfrac{3}{16}<\dfrac{3}{15}=\dfrac{1}{5}\\\Rightarrow S<\dfrac{1}{5}$ Bình luận
$S=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}+\cdots+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\\ 3S=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+\cdots+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ S+3S=1-\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{3}{3^2}-\dfrac{2}{3^2}\right)-\left(\dfrac{4}{3^3}-\dfrac{3}{3^3}\right)+\cdots-\left(\dfrac{100}{3^{99}}-\dfrac{99}{3^{99}}\right)-\dfrac{100}{3^{100}}\\ \Leftrightarrow 4S=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\cdots-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\\ 12S=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-\cdots-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\\ 4S+12S=3-\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Leftrightarrow 16S=3-\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}<3\\ \Rightarrow S<\dfrac{3}{16}<\dfrac{3}{15}=\dfrac{1}{5}\\\Rightarrow S<\dfrac{1}{5}$