Chứng minh B = 1/2^2+1/3^2+1/4^2+...+1/100^2 < 3/4
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a>
\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000
ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )
1/100^2<1/2
=>A<1
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a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/2² + 1/3² + 1/4² + 1/5² + ... + 1/100²
=> A < 1/2.3 + 1/3.4+ 1/4.5 + 1/5.6 + ... + 1/100.101
<=> A < 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/100 - 1/101
<=> A < 1/2 - 1/101
<=> A < 99/202 < 150/202 < 151,5/202
<=> A < 3/4 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{100}\right)\)
\(=0+\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
100 - (1 + 1/2 + 1/3 + 1/4 + ... + 1/100)
= (1 + 1 + 1 + 1 + ... + 1) - (1 + 1/2 + 1/3 + 1/4 + ... + 1/100)
100 số 1 100 phân số
= (1 - 1) + (1 - 1/2) + (1 - 1/3) + (1 - 1/4) + ... + (1 - 1/100)
= 1/2 + 2/3 + 3/4 + ... + 99/100 ( đpcm)
\(B=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}
A = 1/2² + 1/3² + 1/4² + 1/5² + ... + 1/100²
=> A < 1/2.3 + 1/3.4+ 1/4.5 + 1/5.6 + ... + 1/100.101
<=> A < 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/100 - 1/101
<=> A < 1/2 - 1/101
<=> A < 99/202 < 150/202 < 151,5/202
<=> A < 3/4 (đpcm)