chứng minh rằng:
\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
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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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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)\(\dfrac{1}{5^2}+\dfrac{1}{6^5}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
\(S=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\) Chứng minh rằng \(S< \dfrac{1}{2}\)
\(S=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A< 1-\dfrac{1}{50}\Rightarrow A< 1\)
Ta có \(S=\dfrac{1}{2^2}\left(1+A\right)\)
Ta có
\(A< 1\Rightarrow1+A< 2\Rightarrow S< \dfrac{1}{2^2}.2=\dfrac{1}{2}\)
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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:\(\dfrac{1}{3^2}\)+ \(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\) <\(\dfrac{1}{2}\)
Mình cần gấp!
A=1/3^2+1/4^2+1/5^2+1/6^2+...+1/100^2<1/2-1/3+1/3-1/4+...+1/99-1/100
=>A<1/2-1/100<1/2
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25 tháng 7 2021 lúc 15:00
chứng minh rằng :
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Đặt \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
Ta có: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\) (đpcm)
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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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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 , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
Chứng minh rằng : \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4}\)