Gọi biểu thức trên là A

3A = 1 + 1/3 + 1/3^2 + … + 1/3^98`

3A – A = ( 1 + 1/3 + 1/3^2 + … + 1/3^98 ) – ( 1/3 + 1/3^2 + 1/3^3 + … + 1/3^99 )`

2A = 1 – 1/3^99

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

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

\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(\Rightarrow A-\frac{1}{3}A=\left(\frac{1}{3^2}-\frac{1}{3^3}\right)+\left(\frac{1}{3^3}-\frac{1}{3^3}\right)+...+\left(\frac{1}{3}-\frac{1}{3^{100}}\right)\)

\(\Rightarrow\frac{2}{3}A=\frac{1}{3}-\frac{1}{3^{100}}< \frac{1}{3}.\)

\(\Rightarrow A< \frac{1}{3}:\frac{2}{3}\)

\(\Rightarrow A< \frac{1}{2}\left(đpcm\right)\)

Vậy \(A< \frac{1}{2}.\)

Chúc bạn học tốt!

trả lời nhanh dùm

2:

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

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

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

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

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

M=1/3+1/3^2+...+1/3^99

3M=1+1/3+1/3^2+...+1/3^98

3M+1/3^99=1+1/3+...+1/3^99=1+M

3M-M=1-1/3^99

2M=1-1/3^99

M=(1-1/3^99)/2 

Vì 1-1/3^99 <1 nên (1-1/3^99)/2<1/2

Vậy M<1/2