Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
A = \dfrac{1}{3} + \dfrac{2}{{{3^2}}} + \dfrac{3}{{{3^3}}} + \dfrac{4}{{{3^4}}} + ..... + \dfrac{{100}}{{{3^{100}}}}\\
 \Rightarrow 3A = 3.\dfrac{1}{3} + 3.\dfrac{2}{{{3^2}}} + 3.\dfrac{3}{{{3^3}}} + 3.\dfrac{4}{{{3^4}}} + ..... + 3.\dfrac{{100}}{{{3^{100}}}}\\
 \Leftrightarrow 3A = 1 + \dfrac{2}{3} + \dfrac{3}{{{3^2}}} + \dfrac{4}{{{3^3}}} + ...... + \dfrac{{100}}{{{3^{99}}}}\\
 \Rightarrow 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)\\
 \Leftrightarrow 2A = 1 + \left( {\dfrac{2}{3} - \dfrac{1}{3}} \right) + \left( {\dfrac{3}{{{3^2}}} - \dfrac{2}{{{3^2}}}} \right) + \left( {\dfrac{4}{{{3^3}}} - \dfrac{3}{{{3^3}}}} \right) + .... + \left( {\dfrac{{100}}{{{3^{99}}}} - \dfrac{{99}}{{{3^{99}}}}} \right) - \dfrac{{100}}{{{3^{100}}}}\\
 \Leftrightarrow 2A = 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + ..... + \dfrac{1}{{{3^{99}}}} - \dfrac{{100}}{{{3^{100}}}}\\
B = 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + ..... + \dfrac{1}{{{3^{99}}}}\\
 \Leftrightarrow 3B = 3 + 3.\dfrac{1}{3} + 3.\dfrac{1}{{{3^2}}} + 3.\dfrac{1}{{{3^3}}} + 3.\dfrac{1}{{{3^4}}} + .... + 3.\dfrac{1}{{{3^{99}}}}\\
 \Leftrightarrow 3B = 3 + 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + .... + \dfrac{1}{{{3^{98}}}}\\
 \Rightarrow 3B - B = \left( {3 + 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + .... + \dfrac{1}{{{3^{98}}}}} \right) - \left( {1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + ..... + \dfrac{1}{{{3^{99}}}}} \right)\\
 \Leftrightarrow 2B = 3 - \dfrac{1}{{{3^{99}}}}\\
 \Rightarrow 2A = B - \dfrac{{100}}{{{3^{100}}}} < B\\
 \Rightarrow A = \dfrac{B}{2} = \dfrac{{3 - \dfrac{1}{{{3^{99}}}}}}{4} = \dfrac{3}{4} - \dfrac{1}{{{{4.3}^{99}}}} < \dfrac{3}{4}
\end{array}\)

A=$\frac{1}{3}$+$\frac{2}{3^{2}}$+$\frac{3}{3^{3}}$+$\frac{4}{3^{4}}$...+$\frac{100}{3^{100}}$

3A=1+$\frac{2}{3}$+$\frac{3}{3^{2}}$+$\frac{4}{3^{3}}$+...+$\frac{100}{3^{99}}$

3A-2A=1+$\frac{-1}{3^{2}}$+$\frac{-2}{3^{3}}$+...+$\frac{-98}{3^{99}}$-$\frac{200}{3^{100}}$

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

A=1-A-$\frac{100}{3^{100}}$

A+A=1-$\frac{100}{3^{100}}$

2A=1-$\frac{100}{3^{100}}$

A=$\frac{1-\frac{100}{3^{100}}}{2}$ 

A=$\frac{2}{4}$ -$\frac{50}{3^{100}}$<$\frac{3}{4}$

Vậy $\frac{1}{3}$+$\frac{2}{3^{2}}$+$\frac{3}{3^{3}}$+$\frac{4}{3^{4}}$...+$\frac{100}{3^{100}}$<$\frac{3}{4}$(ĐPCM)

Cho mik xin TRẢ LỜI HAY NHẤT NHÉ!!!