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Xét vế trái: A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
=>\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
=>\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
=>\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
=>\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=VP\)
=>đpcm (VP là vế phải)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
http://olm.vn/hoi-dap/question/126681.html
Bạn tham khảo nhé
Trước hết ta biến đổi A thành \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Do đó : \(A=\left[\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right]+\left[\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right]\)
Ta có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75},\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)nên
\(A>\frac{1}{75}\cdot25+\frac{1}{100}\cdot25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(A< \frac{1}{51}\cdot25+\frac{1}{76}\cdot25< \frac{1}{50}\cdot25+\frac{1}{75}\cdot25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Vậy \(\frac{7}{12}< A< \frac{5}{6}\)