CHO MÌNH HỎI:
CHO K = (1/2^2-1)*(1/3^2-1)*(1/4^2-1)*...*(1/100^2-1). SO SÁNH K VỚI -1/2
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\(K=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9999}{10000}=\left(-1\right)^{99}.\frac{1.3.2.4...99.101}{2.2.3.3.4.4...100.100}=-\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}=-\frac{1}{100}.\frac{101}{2}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
Ta co:\(A=\frac{1}{2.2}+\frac{1}{4.4}+\frac{1}{6.6}+...+\frac{1}{14.14}< \frac{2}{2.4}+\frac{2}{4.6}+\frac{1}{6.8}+...+\frac{2}{14.16}\left(1\right)\)
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{14.16}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{16}\)
\(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}< \frac{8}{16}=\frac{1}{2}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow A< \frac{1}{2}\)
V...
Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)
\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)
\(=2-\frac{1}{5}\)< 2
Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2