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= (1/102+1/202) + (1/103+1/201) + ... + (1/151+1/153) + 1/152 / 2/102.202 + ... + 2/151.153 + 2/152.152
= 304/102.202 + 304/103.201 + ... + 304/151.153 + 152/152.152 / 2/102.202 + ... + 2/151.153 + 2/152.152
= 152 (2/102.202 + 2/103.201 + ... + 2/151.153 + 1/152.152) / 2 (1/102.202 + 1/103.201 + ... + 1/152.152)
= 152 / 2
= 76
Ta có: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(1)
Thay (1) vào đề bài
\(\Rightarrow\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}=1\)
\(\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}\)
\(=1\)
Chứng minh:\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{149}+\frac{1}{150}>\frac{1}{3}\)
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\(A>\left(\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}\right)+\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)\) (mỗi ngoặc có 50 số hạng)
\(;A>\left(\frac{1}{150}.50\right)+\left(\frac{1}{200}.50\right)=50.\left(\frac{1}{150}+\frac{1}{200}\right)=50.\frac{7}{600}=\frac{7}{12}\)
