Cho B = \(\dfrac{1}{100^2}+\dfrac{1}{101^2}+....+\dfrac{1}{199^2}\)
Chứng minh:
\(\dfrac{1}{100}< B< \dfrac{1}{99}\)
Những câu hỏi liên quan
Chứng minh : \(\dfrac{1}{2}< \dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+......................+\dfrac{1}{198}+\dfrac{1}{199}+\dfrac{1}{200}< \dfrac{100}{101}\)
Ta có:\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)
Lại có:
\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{100}{101}\)
Vậy ...
Những dãy trên đều có 100 số hạng.
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\(\dfrac{1}{200}< \dfrac{1}{100^2}+\dfrac{1}{101^2}+...+\dfrac{1}{199^2}< \dfrac{1}{99}\)
1/ Cho A dfrac{1}{3}-dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+.....+dfrac{99}{3^{99}}-dfrac{100}{3^{100}} Chứng minh A dfrac{3}{16}2/ Cho B(dfrac{1}{2^2}-1)(dfrac{1}{3^2}-1)....(dfrac{1}{100^2}-1) So sánh B và dfrac{-1}{2}
Đọc tiếp
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
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Cho A = \(\dfrac{1}{100^2}+\dfrac{1}{101^2}+...+\dfrac{1}{199^2}\) . CM \(\dfrac{1}{200}< A< \dfrac{1}{99}\)
Chứng minh rằng số tự nhiên A chia hết cho 101 với:
A=1.2.3...99.100,(1\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}\))
Ta có: A=1.2.3.....99.100.(\(1+\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{99}+\dfrac{1}{100}\))
\(=1.2.3...100\left[\left(1+\dfrac{1}{100}\right)+\left(\dfrac{1}{2}+\dfrac{1}{99}\right)+......+\left(\dfrac{1}{50}+\dfrac{1}{51}\right)\right]\)
=>A= 1.2...100.\(\left[\dfrac{101}{100}+\dfrac{101}{2.99}+......+\dfrac{101}{50.51}\right]\)
=1.2.....100.101\(\left[\dfrac{1}{100}+\dfrac{1}{2.99}+.....+\dfrac{1}{50.51}\right]⋮101\)
Vậy A chia hết cho 101
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18 tháng 3 2023 lúc 19:38
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
25 tháng 7 2021 lúc 11:49
Chứng minh : \(\dfrac{99}{100}\) > \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{99}{202}\)
Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2},\dfrac{1}{3^2}< \dfrac{1}{2.3},...,\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(A\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A<\(1-\dfrac{1}{100}=\dfrac{99}{100}\)(đpcm)
Ta có: \(\dfrac{1}{2^2}>\dfrac{1}{2.3},\dfrac{1}{3^2}>\dfrac{1}{3.4},...,\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
A>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)
A>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
A>\(\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}\)(đpcm)
Vậy \(\dfrac{99}{100}>A>\dfrac{99}{202}\)
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Bài 40: Chứng minh rằng:
a) \(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=9\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}=\dfrac{9}{10}\)
Tính \(\left(100+\dfrac{99}{2}+\dfrac{98}{3}+... +\dfrac{2}{99}+\dfrac{1}{100}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}\right)-2\)