\(B=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(=9.\left(\frac{1}{3}-\frac{1}{675}\right)\)
\(=9.\frac{224}{675}\)
\(=\frac{224}{75}\)
THIẾU
\(=\frac{224}{75}< \frac{225}{75}=3\)
Vậy \(B< 3\)
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