Đặt A = 1/3 + 2/3² + 3/3³ + 4/3^4 + ... + 100/3^100 

=> 3A= 1 + 2/3 + 3/3² + 4/3³ + .... + 100/3^99 

=> 3A-A = 1 + (2/3 - 1/3) + (3/3² - 2/3²) +...+ (100/3^99 - 99/3^99) - 100/3^100

=> 2A= 1+ 1/3 + 1/3² + 1/3³ +...+ 1/3^99 - 100/3^100

Đặt B = 1/3 + 1/3² + 1/3³ +...+ 1/3^99 

=> 3B = 1 + 1/3 + 1/3² + 1/3³ +...+ 1/3^98

=> 2B = 1 - 1/3^99 => B = (1 - 1/3^99)/2

Thay vào 2A => 2A= 1+ 1/2 - 1/(2x3^99) - 100/3^100 < 1+ 1/2 = 3/2 

=> A < 3/4

....

Ý Trước

Đặt \(A=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(3A=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

=>\(3A+A=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}-\frac{3}{3^3}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(4A=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(B=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>\(3B=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>\(3B+B=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>\(4B=-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(B=\frac{-3^{99}-1}{4\cdot3^{99}}\)

Ta có: \(4A=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+B-\frac{100}{3^{100}}=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1+\frac{-1}{4}-\frac{403}{4\cdot3^{100}}\)

=>\(4A<\frac34\)

=>\(A<\frac{3}{16}\)

mà \(\frac{3}{16}<\frac{4}{16}=\frac14\)

nên \(A<\frac14\)

Túi ko bt

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

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

\(\rightarrow2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

6A = \(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\rightarrow4A=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)

4A = \(3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+ \dfrac{100}{3^{100}}\)

4A = 3 - \(\dfrac{203}{3^{100}}\) < 3

\(\Rightarrow\) A < \(\dfrac{3}{4}\) ( đpcm )