Ta có
\(A=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)
\(A=\frac{2}{1.2}-\frac{1}{1.2}+\frac{4}{3.4}-\frac{3}{3.4}+\frac{6}{5.6}-\frac{5}{5.6}+...+\frac{100}{99.100}-\frac{99}{100.99}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
<=>A=1-1/100=99/100
=>7/12<A<5/6(bấm máy tính là biết)
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+\frac{1}{99\cdot100}.CM:\frac{7}{12}< A< \frac{5}{6}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
CMR 7/12<A<5/6
cho A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....+\frac{1}{90\cdot100}\)
chứng minh rằng: 7/20<A<5/16
Cho A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+......+\frac{1}{99\cdot100}\)CMR:\(\frac{7}{12}\)<A<\(\frac{5}{6}\)
Chứng minh rằng
a, B = \(\frac{1\cdot2-1}{2!}+\frac{2\cdot3-1}{3!}+\frac{3\cdot4-1}{4!}+....+\frac{99\cdot100-1}{100!}< 2\)
c, C = \(\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{19}{9^2\cdot10^2}< 1\)
Cho \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....+\frac{1}{99\cdot100}\)
\(B=\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+.....+\frac{1}{100}\)
Khi đó A-b=????
Cho A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+........+\frac{1}{99\cdot100}\)và B=\(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+.......+\frac{1}{100}\)khi đó A-B =.......? ai giải thích hộ mk vs nhá. thanks
CMR A<\(\frac{5}{6}\)voi A=\(\frac{1}{1\cdot2}\)+\(\frac{1}{3\cdot4}\)+\(\frac{1}{5\cdot6}\)+...+\(\frac{1}{99\cdot100}\)
CMR A<\(\frac{5}{6}\)voi A=\(\frac{1}{1\cdot2}\)+\(\frac{1}{3\cdot4}\)+\(\frac{1}{5\cdot6}\)+...+\(\frac{1}{99\cdot100}\)
