CM: \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
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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 : \(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+......+\frac{1}{100^2}<\frac{1}{4}\)
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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Cho A=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+........+\frac{1}{100^2}\)
CM A<\(\frac{3}{2}\)
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= \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{99}{100!}< 1\)
\(\frac{99}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}< \frac{99}{202}CM\)
Chứng minh rằng:
a/\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
b/\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\frac{3}{4}\)
c/\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Bài 1: Chứng minh rằng:1)frac{1}{5} Afrac{1}{5^2}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{2007^2}2)Bfrac{1}{4}+frac{1}{9}+frac{1}{16}+...+frac{1}{81}+frac{1}{100}frac{65}{132}3)Cfrac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{100^2} frac{3}{4}4)frac{1}{6} Dfrac{1}{5^2}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{100^2}5)Efrac{1}{4^2}+frac{1}{6^2}+frac{1}{8^2}+...+frac{1}{100^2} frac{1}{4}Bài 2 : Cho Dfrac{12}{left(2cdot4right)^2}+frac{20}{left(4cdot6right)^2}+...+frac{388}{left(96cdot98right)^2}+frac...
Đọc tiếp
Bài 1: Chứng minh rằng:
1)\(\frac{1}{5}< A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}\)
2)\(B=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}>\frac{65}{132}\)
3)\(C=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{3}{4}\)
4)\(\frac{1}{6}< D=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
5)\(E=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Bài 2 : Cho \(D=\frac{12}{\left(2\cdot4\right)^2}+\frac{20}{\left(4\cdot6\right)^2}+...+\frac{388}{\left(96\cdot98\right)^2}+\frac{396}{\left(98\cdot100\right)^2}\)
Hãy so sánh\(D\) với \(\frac{1}{4}\)
Cảm ơn các bạn nhiều!