Cho: A=1+1/2+1/3+1/4+...+1/2100-1, chứng minh rằng A>50 và A<100
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Khách
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QD
So sánh P = \(\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+...+\dfrac{1}{10^2+11^2}\) với \(\dfrac{9}{20}\)
0
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QD
So sánh P = \(\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+...+\dfrac{1}{10^2+11^2}\) với \(\dfrac{9}{20}\)
0
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QD
So sánh P = \(\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+...+\dfrac{1}{10^2+11^2}\) với \(\dfrac{9}{20}\)
0
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NT
0
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PM
0
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PM
1
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MH
27 tháng 3 2022
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+...+\dfrac{2}{99.100}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{2.99}{100}=\dfrac{99}{50}\)