\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
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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!
Chứng minh frac{1}{5}+frac{1}{13}+frac{1}{14}+frac{1}{15}+frac{1}{61}+frac{1}{62}+frac{1}{63} frac{1}{2}frac{1}{2}b,frac{1}{^{2^2}}+frac{1}{4^2}+frac{1}{6^2}+...+frac{1}{100^2} frac{1}{2}c,frac{3}{4}+frac{8}{9}+frac{15}{16}+...+frac{2499}{2500}48d,frac{1}{6} frac{1}{5^2}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{100^2} frac{1}{4}
Đọc tiếp
Chứng minh \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)\(\frac{1}{2}\)
b,\(\frac{1}{^{2^2}}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
c,\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
d,\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
a, \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Ta có: \(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
Vậy...
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b, Đặt A là tên của tổng trên
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Đặt B là biêu thức trong ngoặc
Ta có: \(1=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow B< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 2-\frac{1}{50}< 2\)
Thay B vào A ta được:
\(A< \frac{1}{2^2}.2=\frac{1}{2}\)
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c, Đặt C = \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
\(C=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(C=\left(1+1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
\(C=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
Đặt D là biểu thức trong ngoặc
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow D< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow-D>-1\)
=>\(C=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>49-1=48\)
Vậy C > 48
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Xem thêm câu trả lời
Bài 1 : Chứng minh
a) \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
b) \(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}...\frac{9999}{10000}< \frac{1}{100}\)
A=\(1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Đặt B=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+..+\)\(\frac{1}{99.100}=\)\(1-\frac{1}{100}< 1\)
Mà A=1+B=>A=1+B<1+1=2
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\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
vậy \(A=\frac{99}{100}< 2\left(đpcm\right)\)
B)
ta có : \(1=1\)
\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)
\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{7}< \frac{1}{4}+...+\frac{1}{4}=\frac{4}{4}=1\)
\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+...+\frac{1}{8}=\frac{8}{8}=1\)
\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{63}< 1\)
tất cả công lại \(\Rightarrow B< 6\)
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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 \(\frac{1}{6}< A< \frac{1}{4}\left(đpcm\right)\)
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Chứng minh:
a, M= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\)
b, \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
CMR:N=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
\(\frac{1}{2^2}n=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(\frac{1}{4}n=\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{102}}\)
\(n-\frac{1}{4}n=\frac{1}{2^2}-\frac{1}{2^{102}}\)
\(\frac{3}{4}n=\frac{1}{4}-\frac{1}{2^{102}}< \frac{1}{4}\)
\(n< \frac{1}{4}.\frac{3}{4}\) suy ra n>1/2
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Chứng minh : \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+......+\frac{1}{100^2}< \frac{1}{4}\)\(\frac{1}{4}\)
ok, ta co \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
\(A< \frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)
\(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+..+\frac{1}{99}-\frac{1}{100}\)
\(A< \frac{1}{4}-\frac{1}{100}\)
\(A< \frac{1}{4}\)
Lai co \(A>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{100\cdot101}=\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+..+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}\)
\(A>\frac{1}{6}\)
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\(A=\left(2+4+6+...+100\right)\left(\frac{3}{5}:0,7+3\left(\frac{-2}{7}\right)\right):\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
A=[2+4+6+...+100][3/5:0,7+3[-2/7]]:[1/2+1/4+1/6+...+1/100]
A=[2+4+6+...+100][6/7+[-6/7]]:[1/2+1/4+1/6+...+1/100]
A=[2+4+6+...+100][0]:[1/2+14+1/6+...+1/100]
A=0
CHỈ MK CÁCH VIẾT PHÂN SỐ ĐI
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chứng minh rằng\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+.......+\frac{1}{100^2}< \frac{1}{4}\)