Cho A =\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^4}\)+\(\dfrac{1}{2^6}\)+....+\(\dfrac{1}{2^{100}}\).C/minh A <\(\dfrac{1}{3}\)
Những câu hỏi liên quan
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}
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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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25 tháng 7 2021 lúc 11:54
chứng minh :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}\) b) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
25 tháng 7 2021 lúc 14:43
chứng minh rằng :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\) b)\(\dfrac{1}{5^2}+\dfrac{1}{6^5}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
cho A =\(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)
Chứng minh rằng A<\(\dfrac{1}{3}\)
Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow2^2A=1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{98}}\)
\(\Rightarrow2^2A-A=\left(1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\right)\)
\(\Rightarrow3A=1-\dfrac{1}{2^{100}}\)
\(\Rightarrow A=\dfrac{1-\dfrac{1}{2^{100}}}{3}< \dfrac{1}{3}\)(đpcm)
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Cho A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+....+\(\dfrac{1}{100^2}\). Chứng minh rằng A<\(\dfrac{3}{4}\)
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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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Cho A=1+\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2^{100}-1}\)
Chứng minh rằng 50<A<100
28 tháng 4 2022 lúc 12:32
a)chứng minh rằng :dfrac{1}{3^2}+dfrac{1}{4^2}+dfrac{1}{5^2}+dfrac{1}{6^2}........+dfrac{1}{100^2} dfrac{1}{2}b)tính nhanh tổng S với S dfrac{1}{3.5}+dfrac{1}{5.7}+dfrac{1}{7.9}+......+dfrac{1}{61.63}các cao nhân gải giúp với ạ !!! iem đang cần gấp
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a)chứng minh rằng :\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)........+\(\dfrac{1}{100^2}< \dfrac{1}{2}\)
b)tính nhanh tổng S với S= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+......+\dfrac{1}{61.63}\)
các cao nhân gải giúp với ạ !!! iem đang cần gấp
Cho: A= 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2^{100}-1}\)
Chứng minh rằng: 50 < A < 100
Giúp mình với!