2:

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

=>\(B=\frac{3^{99}-1}{2\cdot3^{99}}\)

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

\(=1+\frac{3^{99}-1}{2\cdot3^{99}}-\frac{100}{3^{100}}=\frac{2\cdot3^{100}+3^{100}-3-100}{3^{100}\cdot2}=\frac{3\cdot3^{100}-103}{2\cdot3^{100}}=\frac32-\frac{103}{2\cdot3^{100}}<\frac32\)

=>\(A<\frac34\)

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

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

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

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

Đúng rồi đó ngu còn bày đặt

xin lỗi nha mình gửi lộn ak

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

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

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

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

\(6B=3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(6B-2B=\left(3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)\)

\(4B=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)

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

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

\(\Rightarrow C< \dfrac{3}{4}\left(đpcm\right)\)

nhân 3 trừ đi sau đó xét cái sau sẽ thấy B<3/4

Lười lắm

\(B=\dfrac{1}{3}+\dfrac{2}{3^2}+...+\dfrac{100}{3^{100}}\\ \Rightarrow3B=1+\dfrac{2}{3}+....+\dfrac{100}{3^{99}}\\ \Rightarrow2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\\ A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^{99}}\\ \Rightarrow3A=3+1+..+\dfrac{1}{3^{98}}\\ \Rightarrow2.A=3-\dfrac{1}{3^{99}}< 3\\ \Rightarrow A< \dfrac{3}{2}\\ \Rightarrow2B=A-\dfrac{100}{3^{100}}< \dfrac{3}{2}\\ \Rightarrow B< \dfrac{3}{4}\)

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

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

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

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

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

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

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

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

=> 4M < \(\dfrac{3}{4}\) => M < \(\dfrac{3}{16}\) ĐPCM

Đặt :

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

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

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

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

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

Đặt :

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

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

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

\(4B=3-\dfrac{1}{99}< 3\Rightarrow B< \dfrac{3}{4}\)

\(\Rightarrow4A< \dfrac{3}{4}\Rightarrow A< \dfrac{3}{16}\rightarrowđpcm\)