Cho: S=\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2018^2}\)chứng tỏ S< \(\dfrac{1}{4}\)
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Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)
Cho \(S=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\). Chứng tỏ rằng S<\(\dfrac{1}{16}\)
23 tháng 4 2023 lúc 15:20
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\) . Chứng tỏ rằng \(S< \dfrac{1}{16}\)
22 tháng 2 2022 lúc 16:48
a, Tính: M = \(1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
b, Chứng tỏ: S = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b: 
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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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Chứng tỏ rằng B = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}< 1\)
Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...............
\(\dfrac{1}{8^2}< \dfrac{1}{7.8}\)
=> B < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{7.8}\)
B < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
B < \(1-\dfrac{1}{8}< 1\) (Do \(\dfrac{1}{8}>0\))
Vậy.....
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cho \(S=\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\)
a, CMR : \(S< \dfrac{1}{24}\)
b, Có hay ko số nguyên x thỏa mãn : \(3x^7-2x^4+6x^2-18x=501\)
9 tháng 5 2022 lúc 20:03
Cho S= \(\dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\).Chứng tỏ rằng S<2
\(S=2S-S\)
\(=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^{2022}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^{2022}}\right)\)
\(=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^{2022}}\right)\)
\(=2-\dfrac{1}{2^{2022}}< 2\left(đpcm\right)\)
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Cho S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2015^2}\). Chứng tỏ rằng \(\dfrac{1007}{2016}< S< \dfrac{2014}{2015}\)
Lời giải:
Ta có:
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\)
\(S> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}\)
\(\Leftrightarrow S> \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2016-2015}{2015.2016}\)
\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{2016}=\frac{1007}{2016}\)
--------------------------
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}\)
\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2014}{2015}\)
\(\Leftrightarrow S< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2015-2014}{2014.2015}\)
\(\Leftrightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2014}-\frac{1}{2015}\)
\(\Leftrightarrow S< 1-\frac{1}{2015}=\frac{2014}{2015}\)
Vậy ta có đpcm.
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