\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow3A+A=\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
-Đặt \(S=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)
\(3S=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}\)
\(\Rightarrow3S+S=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow4S=3-\dfrac{1}{3^{99}}=\dfrac{3^{100}-1}{3^{99}}\)
\(\Rightarrow S=\dfrac{3^{100}-1}{4.3^{99}}\)
\(\Rightarrow4A=\dfrac{3^{100}-1}{4.3^{99}}-\dfrac{100}{3^{100}}=\dfrac{3\left(3^{100}-1\right)-400}{4.3^{100}}=\dfrac{3^{101}-403}{4.3^{100}}\)
\(\Rightarrow A=\dfrac{3^{101}-403}{16.3^{100}}< \dfrac{3^{101}}{16.3^{100}}=\dfrac{3}{16}\)
