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 \(A< \frac{1}{2}.\)

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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

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

=>3S=1+1/3+1/3^2+1/3^3+....+1/3^98

=>3S-S=(1+1/3+1/3^2+...1/3^98)-(1/3+1/3^2+...+1/3^99)

=>2S=1-1/3^99

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

=>S=(3^99-1)/2.3^99

=>S=1/2-1/2.3^99.

Vì 1/2-1/2.3^99<1/2

=>S<1/2 (đpcm)

Ta có:1/(3^n)+1/(3^(n+1))=2/(3^(n+1)

Áp dụng ta có:1-1/3=2/3 

1/3-1/(3^2)=2/(3^2) 

1/(3^2)-1/(3^3)=2/(3^3) 

1/(3^98)-1/(3^99)=2/(3^99). 

Cộng từng vế các phép tính với nhau ta có:1-1/(3^99)=2M. 

Mà 1-1/(3^99)<1 nên 2M<1 nên M<1/2(điều phải chứng minh)