Sửa đề: Chứng minh B<1
Ta có: \(B=\frac{1}{1+3}+\frac{1}{1+3+5}+\cdots+\frac{1}{1+3+5+\cdots+101}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{51^2}\)
TA có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{51^2}<\frac{1}{50\cdot51}=\frac{1}{50}-\frac{1}{51}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{51^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{50}-\frac{1}{51}\)
=>\(A<1-\frac{1}{51}\)
=>A<1
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