\(ChoA=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+....+\frac{1}{99\times100}\)\
Chứng minh rằng
5/6 < A <7/12
Những câu hỏi liên quan
Chứng minh rằng :\(\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+...+\frac{1}{99\times100}=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
cảm ơn bạn nha
\(A=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+...+\frac{1}{99\times100}.CMR:\frac{7}{12}< A< \frac{5}{6}\)
Cho \(A=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+...+\frac{1}{99\times100}\)
Chứng minh rằng \(\frac{7}{12}\)<A<\(\frac{5}{6}\)
A=1/1.2+1/12+...+1/99.100
A=7/12+...1/99.100
Suy ra A>7/12 (1)
A=1-1/2+1/3-1/4+...+1/99-1/100
A=(1/2+1/3)-(1/4-...+1/100)
A=5/6-(1/4-...+1/100)
suy ra A<5/6 (2)
Vậy 7/12<A<5/6
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Lê Tùng lâm bài của bạn chưa đúng vì
A = \(\frac{1}{1.2}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
Chứ không phải là: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{98.99}+\frac{1}{99.100}\)
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Chứng minh rằng:
\(\frac{7}{12}\)<\(\frac{1}{1\times2}\)+\(\frac{1}{3\times4}\)+\(\frac{1}{5\times6}\)+ ..... +\(\frac{1}{99\times100}\)<\(\frac{5}{6}\)
\(cmr;\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+.....+\frac{1}{99\times100}=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+.....+\frac{1}{100}\)
ai làm đung mình tick cho
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
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\(\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\div\left(\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+...+\frac{1}{99\times100}\right)\)
Cho A =\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{99\times100}\)
Chứng minh rằng \(\frac{5}{6}\)< A < \(\frac{7}{12}\)
Chứng minh rằng: \(\frac{1\times2-1}{2!}+\frac{2\times3-1}{3!}+\frac{3\times4-1}{4!}+...+\frac{99\times100-1}{100!}<2\)
1\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+............+\frac{1}{99\times100}+\frac{1}{100\times101}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}.\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
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