Lời giải:

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

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

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

\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+....-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow 16A=12A+4A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}<3\)

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

Lời giải:

$M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}$

$3M=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}$

$\Rightarrow 2M=3M-M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}$

$2M+\frac{100}{3^{100}}=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}$

$3(2M+\frac{100}{3^{100}})=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}

$\Rightarrow 2(2M+\frac{100}{3^{100}})=3(2M+\frac{100}{3^{100}})-(2M+\frac{100}{3^{100}})=2-\frac{1}{3^{99}}$

$M=\frac{1}{2}-\frac{1}{4.3^{99}}-\frac{50}{3^{100}}<\frac{1}{2}< \frac{3}{4}$ 
Ta có đpcm.

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

\(\Rightarrow3M=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)

\(\Rightarrow3M-M=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+...+\dfrac{100}{3^{100}}\right)\)

\(\Rightarrow2M=1+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\dfrac{100}{3^{100}}\)

\(\Rightarrow2M=1+\dfrac{1}{2}-\dfrac{1}{3^{99}.2}-\dfrac{100}{3^{100}}\)

\(\Rightarrow M=\dfrac{3}{4}-\dfrac{1}{3^{99}.4}-\dfrac{50}{3^{100}}< \dfrac{3}{4}\)

Vậy...

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

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

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

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

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

Đặt:

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

\(3B=3\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)

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

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

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

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

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

Ta có đpcm

2:

\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)

xin lỗi nha mình gửi lộn ak