\(\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{2}{2020^2}\)
Chứng minh dãy phân số trên < \(\dfrac{1}{2}\)
Những câu hỏi liên quan
chứng minh :\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+.............+\dfrac{1}{2020^2}< \dfrac{1}{4}\)
vì: \(\dfrac{1}{4^2}< \dfrac{1}{4}\)
\(\dfrac{1}{6^2}< \dfrac{1}{4}\)
........
\(\dfrac{1}{2020^2}< \dfrac{1}{4}\)
=> \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{2020^2}< \dfrac{1}{4}\)
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\(A=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{2020}}-\dfrac{1}{2^{2022}}
\)
Chứng minh A<0.2
4A=1-1/2^2+1/2^4-...+1/2^2018-1/2^2020
=>5A=1-1/2^2022
=>A=1/5-1/5*2^2022<1/5=0,2
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*Thực hiện1/ (dfrac{2021}{2020}+dfrac{2020}{2021}) x (dfrac{1}{2}-dfrac{1}{3}-dfrac{1}{6})2/ (dfrac{7}{19}-dfrac{5}{12}):dfrac{-5}{8}-(dfrac{7}{19}-dfrac{29}{12}):dfrac{5}{8}3/ dfrac{-5}{6}xdfrac{7}{24}-dfrac{5}{6}xdfrac{14}{24}-dfrac{5}{6}xdfrac{3}{24}
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*Thực hiện
1/ (\(\dfrac{2021}{2020}\)+\(\dfrac{2020}{2021}\)) x (\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\))
2/ (\(\dfrac{7}{19}\)-\(\dfrac{5}{12}\)):\(\dfrac{-5}{8}\)-(\(\dfrac{7}{19}\)-\(\dfrac{29}{12}\)):\(\dfrac{5}{8}\)
3/ \(\dfrac{-5}{6}\)x\(\dfrac{7}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{14}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{3}{24}\)
1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)
=\(0\)
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mink chịu bài này nó rất khó
Chứng minh \(\dfrac{1}{5}\)< \(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)+......+\(\dfrac{1}{99^2}\)+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{3}\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{100\cdot101}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\left(1\right)\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3}\)
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11 tháng 4 2022 lúc 8:21
chứng minh rằng : \(\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)
1/4^2 + 1/5^2 +... + 1/100^2 < 1/3.4 + 1/4.5 +...+ 1/99.100
A=1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100
=1/3 - 1/100 < 1/3
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26 tháng 2 2022 lúc 14:24
Dãy phân số sắp xếp theo thứ tự tăng dần là:
\(A.\dfrac{1}{2};\dfrac{1}{4};\dfrac{1}{6};\dfrac{1}{5}\) \(B.\dfrac{1}{6};\dfrac{1}{5};\dfrac{1}{4};\dfrac{1}{2}\)
\(C.\dfrac{1}{2};\dfrac{1}{6};\dfrac{1}{5};\dfrac{1}{4}\) \(D.\dfrac{1}{2};\dfrac{1}{4};\dfrac{1}{5};\dfrac{1}{6}\)
Xem thêm câu trả lời
Chứng minh rằng \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Ta có: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
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\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
......
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
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Chứng minh rằng dãy số sau đây tăng và bị chặn trên :
\(x_1=\dfrac{1}{5+1};x_2=\dfrac{1}{5+1}+\dfrac{1}{5^2+1};x_3=\dfrac{1}{5+1}+\dfrac{1}{5^2+1}+\dfrac{1}{5^3+1},.....;x_n=\dfrac{1}{5+1}+\dfrac{1}{5^2+1}+.....+\dfrac{1}{5^n+1}\)
6 tháng 3 2021 lúc 12:52
Cho A = \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+.....+\dfrac{1}{2019^2}\)
Chứng minh rằng \(\dfrac{20}{101}< A< \dfrac{1}{4}\)
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{2019\cdot2020}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=\dfrac{1}{5}-\dfrac{1}{2020}=\dfrac{404-1}{2020}=\dfrac{403}{2020}>\dfrac{40}{2020}=\dfrac{20}{101}\left(1\right)\) \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}< \dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{2018\cdot2019}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2018}-\dfrac{1}{2019}=\dfrac{1}{4}-\dfrac{1}{2019}=\dfrac{2019-4}{4\cdot2019}=\dfrac{2015}{4\cdot2019}< \dfrac{2019}{4\cdot2019}=\dfrac{1}{4}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{20}{101}< A< \dfrac{1}{4}\)
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