CM: \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Những câu hỏi liên quan
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
Đúng 0
Bình luận (0)
CM
\(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+....+\frac{1}{100^2}\)
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 \(\frac{7}{12}< A< \frac{5}{6}\)
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!
CMR : \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...\frac{1}{100^2}\)
Ta có :
\(A< \frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+...+\frac{1}{99\times100}\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
Ta có :
\(A>\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{100\times101}\)
\(\Leftrightarrow A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{100}>\frac{1}{6}\)
Vậy \(\frac{1}{6}< A< \frac{1}{4}\left(đpcm\right)\)
Đúng 0
Bình luận (0)
chứng minh rằng\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+.......+\frac{1}{100^2}< \frac{1}{4}\)
Chứng minh rằng :
\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Chứng minh rằng :
\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
chứng minh rằng:\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+.......+\frac{1}{100^2}< \frac{1}{4}\)
Ta có: \(\frac{1}{5^2}< \frac{1}{4.5};\frac{1}{6^2}< \frac{1}{5.6};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Cộng vế với vế ta được: \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)(1)
Tương tự: \(\frac{1}{5^2}>\frac{1}{5.6};\frac{1}{6^2}>\frac{1}{6.7};...;\frac{1}{100^2}>\frac{1}{100.101}\)
Cộng vế với vế ta được \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)(2)
Từ (1) và (2) =>đpcm