Cho A=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+........+\frac{1}{100^2}\)
CM A<\(\frac{3}{2}\)
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Cho A = \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+..+\frac{1}{2^{100}}\)
CM: A < \(\frac{1}{3}\)
tính ra nhé bạn. nếu bạn ko bít cách tính
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\(2^2.A=\frac{2^2}{2^2}+\frac{1}{2^2}+\frac{1}{2^4}+..+\frac{1}{2^{98}}\)
2^2A-A=3A
\(3A=\frac{2^2}{2^2}-\frac{1}{2^{98}}=1-\frac{1}{2^{98}}<1\)
=> A<1/3=> dpcm
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a) CM: A2= \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{100}}>10\)
b) CM: A3= \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{99}{100!}< 1\)
CM: \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
CM\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(...\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+...+\frac{1}{99.100}\)
Mình chỉ làm được đến đây thôi. Sorry nha. À mà bạn thử vào trang này xem https://vn.answers.yahoo.com/question/index?qid=20121102064330AAkYsXP
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CM \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)<\(\frac{3}{4}\)
A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)CM \(\frac{7}{12}< A< \frac{5}{6}\)
Cho A=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
B=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
CM A<B
b, Cm A <\(\frac{1}{10}\)
Cho A=\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
CMR \(\frac{1}{6}< A< \frac{1}{4}\)
CM : \(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+......+\frac{1}{100^2}<\frac{1}{4}\)
đặt 1/5^2+1/6^2+...+1/100^2=A
ta có: \(A<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+..+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\left(1\right)\)
\(A>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+..+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+..+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\left(do\frac{1}{5}>\frac{1}{6}\right)\left(2\right)\)
từ (1);(2)=>1/6<A<1/4
=>đpcm
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CM: \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
1/5^2 + 1/6^2 + 1/7^2 + ... + 1/100^2
< 1/4×5 + 1/5×6 + 1/6×7 + ... + 1/99×100
< 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/99 - 1/100
< 1/4 - 1/100 < 1/4 ( đpcm)
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