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

\(\Rightarrow3A=1-\frac{2}{3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

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

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

\(\Rightarrow3B=3+1+...+\frac{3}{3^{98}}\)

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

\(B=\frac{3}{2}-\frac{1}{3^{99}.2}\)

Thay B vào 4A ta có:

\(4A=\frac{3}{2}-\frac{1}{3^{99}.2}\)

\(A=\frac{3}{2.4}-\frac{1}{3^{99}.2.4}\)

\(A=\frac{3}{8}-\frac{1}{3^{99}.8}\)

Vì \(\frac{3}{8}>\frac{3}{16}\)

\(\Rightarrow\frac{3}{8}-\frac{1}{3^{99}.8}< \frac{3}{16}\)

Vậy \(A< \frac{3}{16}\)

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

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

\(\Rightarrow4B=3B+B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

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

\(3M=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)

\(\Rightarrow4M=3M+M=3-\frac{1}{3^{99}}\)

\(\Rightarrow M=\frac{3}{4}-\frac{1}{3^{99}\cdot4}\)

\(\Rightarrow4B=M-\frac{100}{3^{100}}=\frac{3}{4}-\frac{1}{3^{99}\cdot4}-\frac{100}{3^{100}}\)

\(\Rightarrow B=\frac{3}{16}-\frac{1}{3^{99}\cdot16}-\frac{100}{3^{100}\cdot4}\) \(\Rightarrow B< \frac{3}{16}\)

a) \(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)

\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)

\(\Rightarrow3A=2A+A=1-\frac{1}{2^6}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{2^6\cdot3}< \frac{1}{3}\) ( đpcm )

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

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

4B<\(1-\frac{1}{3}+\frac{1}{3^2}-.........+\frac{1}{3^{99}}\)

12B<\(3-1+\frac{1}{3}-.........+\frac{1}{3^{98}}\)

12B+4B<\(3-\frac{1}{3^{99}}\)

16B<3

\(\Rightarrow B<\frac{3}{16}\)

\(\Rightarrow\)B

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

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

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

Ta lại có : 3M = 3 -1 +\(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-......+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)

M + 3M = 3 - \(\frac{1}{3^{99}}\)

4M = 3 - 1/3⁹⁹ => M = 3/4 - 1/4.3⁹⁹

Khi đó : 4A = ( 3/4 - 1/4.3⁹⁹) - 1/3⁹⁹

4A = 3/4 - 1/4.3⁹⁹ - 1/3⁹⁹ < 3/4

=> A < 3/4 : 4

=> A < 3/16 (đpcm)