Chứng minh: 1/3+1/3^2+1/3^3+.......+1/3^99 <1/2
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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 = \(\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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!
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2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
Chứng minh rằng:
a) 1/2-1/4+1/8-1/16+1/32-1/64<1/3
b) 1/3-2/3^2+3/3^3-3/3^4+...+99/3^99-100/3^100<3/16
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)