Đặt :

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

\(\Leftrightarrow3S=1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{3}{3^3}+...........+\dfrac{100}{3^{99}}\)

\(\Leftrightarrow3S-S=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+.........+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+..........+\dfrac{100}{3^{99}}\right)\)

\(\Leftrightarrow2S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...........+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

Đặt :

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

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

\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+........+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}\right)\)

\(\Leftrightarrow2A=1-\dfrac{1}{3^{99}}\)

\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)

\(\Leftrightarrow S=1+\dfrac{1}{2}-\dfrac{1}{2.3^{99}}+\dfrac{100}{3^{100}}< 1+\dfrac{1}{2}=\dfrac{3}{2}< \dfrac{3}{4}\)

\(\Leftrightarrow S< \dfrac{3}{4}\left(đpcm\right)\)

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

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

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

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

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

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:

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

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

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

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

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

a: Đặt \(A=\frac12-\frac14+\frac18-\frac{1}{16}+\cdots-\frac{1}{1024}\)

=>\(A=\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{10}}\)

=>\(2A=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots-\frac{1}{2^9}\)

=>\(2A+A=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots-\frac{1}{2^9}+\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{10}}\)

=>\(3A=1-\frac{1}{2^{10}}<1\)

=>\(A<\frac13\)

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

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

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

=>\(4B=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

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

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

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

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

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

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

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}=\frac34-\frac{403}{4\cdot3^{100}}\)

=>\(4B<\frac34\)

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

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

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

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

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

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

     = 3 - \(\dfrac{1}{3^{99}}\)

B = \(\dfrac{3}{4}-\dfrac{1}{3^{99}\cdot4}\)

⇒ 4A = \(\dfrac{3}{4}-\dfrac{1}{3^{99}\cdot4}\) - \(\dfrac{100}{3^{100}}\) 

A = \(\dfrac{3}{16}-\dfrac{1}{3^{99}\cdot4^2}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)

Vậy A < \(\dfrac{3}{16}\)

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

\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

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

\(\Rightarrow4A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\) (1)

\(\Rightarrow12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\) (2)

Cộng vế (1) và (2):

\(\Rightarrow16A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow16A< 3\)

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

Đặt `A` `=` `1/3 - 2/3^2+3/3^3 - 4/3^4+ ... + 99/3^99-100/3^100`
`=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99`
`=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100`
`=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99`
`=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...`
`=>16A=3-101/3^99-100/3^100`
`<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16`
`=> A<3/16`

@Nae

Lời giải:

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

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

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

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

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

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