Chứng tỏ :
\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^{1999}}>1000\)
HH
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CHứng minh rằng 1 + \(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^{1999}}>1000\)
Bài 1:Chứng tỏ rằng:Bdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+dfrac{1}{5^2}+dfrac{1}{6^2}+dfrac{1}{7^2}dfrac{1}{8^2}1Bài 2:Chứng tỏ rằng:Edfrac{3}{4}+dfrac{8}{9}+dfrac{15}{16}+...+dfrac{2499}{2500}1Bài 3:Chứng tỏ rằng:1dfrac{2011}{2020^2+1}+dfrac{2021}{2020^2+2}+dfrac{2021}{2020^3+3}+...+dfrac{2021}{2020^3+2020} 2
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Bài 1:Chứng tỏ rằng:B=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)\(\dfrac{1}{8^2}\)<1
Bài 2:Chứng tỏ rằng:E=\(\dfrac{3}{4}\)+\(\dfrac{8}{9}\)+\(\dfrac{15}{16}\)+...+\(\dfrac{2499}{2500}\)<1
Bài 3:Chứng tỏ rằng:1<\(\dfrac{2011}{2020^2+1}\)+\(\dfrac{2021}{2020^2+2}\)+\(\dfrac{2021}{2020^3+3}\)+...+\(\dfrac{2021}{2020^3+2020}\)< 2
1:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
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Cho A = \(\dfrac{1001}{1000^2+1}\)+\(\dfrac{1001}{1000^2+2}\)+\(\dfrac{1001}{1000^2+3}\)+...+\(\dfrac{1001}{1000^2+1000}\)
Chứng minh rằng 1<\(^{A^2}\)<4
Tổng A có 1000 số hạng.
Vậy
Chúc bạn học tốt.
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Tổng A có 1000 số hạng
A>(1001/1000^2+1000)*1000=1001*1000/1000*(1000+1)=1
A<(1001/1000^2)*1000=1001/1000=1+1/1000<1
Vậy 1<A<2 nên 1<A^2<4
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Bài 1:
a, Cho A dfrac{1}{2^2}+dfrac{1}{4^2}+dfrac{1}{6^2}+....+dfrac{1}{100^2}
Chứng tỏ: A dfrac{1}{2}
b, Cho B dfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+....+dfrac{1}{2^{20}}
Chứng tỏ B 1
c, Cho C dfrac{1}{5}+dfrac{1}{13}+dfrac{1}{14}+dfrac{1}{15}+dfrac{1}{61}+dfrac{1}{62}+dfrac{1}{63}
Chứng tỏ C dfrac{1}{2}
d, Cho D dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{20^2}
Chứng tỏ D 1
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Bài 1:
a, Cho A = \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}\)
Chứng tỏ: A <\(\dfrac{1}{2}\)
b, Cho B = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{20}}\)
Chứng tỏ B < 1
c, Cho C = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
Chứng tỏ C < \(\dfrac{1}{2}\)
d, Cho D = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{20^2}\)
Chứng tỏ D < 1
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
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A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
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!!!
KJ
18 tháng 2 2022 lúc 18:27
Cho F = \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}=\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\). Chứng tỏ \(F< 1\dfrac{3}{4}\)
sửa đề : \(F=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(\dfrac{1}{1^2}< \dfrac{1}{1.2};\dfrac{1}{2^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
Cộng vế với vế
\(\dfrac{1}{1^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)< 7/4
Vậy ta có đpcm
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NX
20 tháng 4 2021 lúc 20:58
chứng tỏ rằng
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
Chứng tỏ
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{2002^2}+\dfrac{1}{2003^2}< 1\)
=> A<1/1.2 + 1/2.3 + ....+ 1/2001.2002 + 1/2002.2003
=> A< 1- 1/2 + 1/2 - 1/3 + .... + 1/2001 - 1/2002 + 1/2002 - 1/2003
=>A< 1 - 1/2003 < 1
=> A< 1
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TQ
25 tháng 4 2023 lúc 9:53
Chứng tỏ rằng:
a)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
b)\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
b\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4
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Tương tự như vậy với câu a\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2
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Chứng tỏ:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}< \dfrac{2}{3}\)