Ta có: \(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\right)\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{100}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(\Rightarrow4D=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D< 3-\frac{203}{3^{100}}< 3\Rightarrow D< \frac{3}{4}\left(ĐPCM\right)\)
