Question

cho D=1/3+2/3²+...+101/101²

CMR D<3/4

Answer 1

\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)

\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\)

\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}=A-\frac{101}{3^{101}}\)

\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)

\(3A=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}\)

\(\Rightarrow2A=3-\frac{1}{3^{100}}\Rightarrow A=\frac{3}{2}-\frac{1}{2.3^{100}}< \frac{3}{2}\)

\(\Rightarrow2D=A-\frac{101}{3^{101}}< A< \frac{3}{2}\Rightarrow D< \frac{3}{4}\)