Question

Chứng minh :

\(\dfrac{1}{3} -\dfrac{2}{3^{2}} + \dfrac{3}{3^{3}}-...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}<\dfrac{3}{16}\)

Answer 1

M = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

3M = \(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+....+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

M+3M = \(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+....+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

4M < \(1-\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

Đặt A = \(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+....+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

3A = \(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+......+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}\)

A+3A=\(3-\dfrac{1}{3^{99}}\)

4A = \(3-\dfrac{1}{3^{99}}< 3=>A< \dfrac{3}{4}\)

=> 4M < \(\dfrac{3}{4}\) => M < \(\dfrac{3}{16}\) ĐPCM

Answer 2

Đặt :

\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-.............+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

\(3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...............+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

\(3A+A=\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...............+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)\)\(+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-...............+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)\)

\(4A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+..............+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

\(4A< 1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+............+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

Đặt :

\(B=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...........+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

\(3B=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+................+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}\)

\(3B+B=3-\dfrac{1}{3^{99}}\)

\(4B=3-\dfrac{1}{99}< 3\Rightarrow B< \dfrac{3}{4}\)

\(\Rightarrow4A< \dfrac{3}{4}\Rightarrow A< \dfrac{3}{16}\rightarrowđpcm\)