Question

[Đặng Quốc Huy]

Chứng minh rằng: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+....................+\frac{100}{3^{100}}< \frac{3}{4}\)

Answer 1

[Nguyễn Thanh Hằng]

Đặt :

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

\(\Leftrightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\)

\(\Leftrightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+....+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+....+\frac{100}{3^{100}}\right)\)

\(\Leftrightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt : \(H=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\) \(\Leftrightarrow2A=H-\frac{100}{3^{100}}\)

\(\Leftrightarrow3H=3+1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)

\(\Leftrightarrow3H-H=\left(4+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)

\(\Leftrightarrow2H=3-\frac{1}{3^{99}}\)

\(\Leftrightarrow H=\frac{3-\frac{1}{99}}{2}\)

\(\Leftrightarrow2A=\frac{3-\frac{1}{3^{99}}}{2}-\frac{100}{3^{100}}\)

\(\Leftrightarrow A=\frac{1-\frac{1}{3^{99}}}{2}-\frac{100}{2.3^{100}}\)

\(\Leftrightarrow A< \frac{3}{4}\left(đpcm\right)\)

Answer 2

[Vũ Minh Tuấn]

Mà sao bạn tức giận thế nhỉ, mọi khi có thể đâu. Khổ thật.

Nguyễn Văn Đạt