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a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)
\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B
=>B/A=1/100
b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)
\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)
\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
=>A/B=25
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{2}+\frac{1}{2^2}+.............+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+...........+\frac{1}{2^{98}}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+.......+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{99}}\)
\(\Leftrightarrow2^{99}.A=2^{99}-1\left(đpcm\right)\)
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Cho A= 1/2 + 2/2^2 + 3/2^3 + .... + 99/2^99 + 100/2^100
So sánh A với 2
Xl mấy bạn mk gõ thiếu bài gốc đây nhá Giải giúp mk với nạ
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\(\Rightarrow A=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(\Leftrightarrow A=\frac{1}{99.100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(\Rightarrow A=\frac{1}{99.100}-1+\frac{1}{99}\)
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A=-1++(-1)+..+-(1) có 50 số -1
=>A=-1x50=-50
B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
B=0+0+0+..+0
B=0
C=2^100-(2^99+2^98+...+1)
C=2^100-(2^100-1)
C=1
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C = 1/3 + 1/3^2 + 1/3^3 + ... =1/3^99
=> C = 1/3^99 = 1/(3^99)
=> C < 1/2 (đpcm)
2A=2^101-2^100+2^98+...+2^3-2^2
3A = 2A + A
3A = 2^101 - 2 ( Cứ tính là ra , âm vs dương triệt tiêu )
A = (2^101-2) :3
B tăng tự
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A = 2100 - 1