A=\(\frac{1}{1}\)*2+\(\frac{1}{3}\)*4+\(\frac{1}{5}\)*6+...+\(\frac{1}{99}\)*100
chứng minh rằng \(\frac{7}{12}\)<A<\(\frac{5}{6}\)
lưu ý không ghi tắt các chữ
K
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13 tháng 8 2018
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Ta có:
\(\frac{1}{51}>\frac{1}{75}\)
\(\frac{1}{52}>\frac{1}{75}\)
......................
\(\frac{1}{75}=\frac{1}{75}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=25.\frac{1}{75}=\frac{1}{3}\)(1)
Ta có:
\(\frac{1}{76}>\frac{1}{100}\)
\(\frac{1}{77}>\frac{1}{100}\)
........................
\(\frac{1}{100}=\frac{1}{100}\)
\(\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=25.\frac{1}{100}=\frac{1}{4}\)(2)
Từ (1) và (2) ta có:
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{7}{12}\)(5)
Ta có:
\(\frac{1}{51}=\frac{1}{51}\)
\(\frac{1}{52}< \frac{1}{51}\)
...................
\(\frac{1}{75}< \frac{1}{51}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=25.\frac{1}{51}< 25.\frac{1}{50}=\frac{1}{2}\)(3)
Ta có:
\(\frac{1}{76}=\frac{1}{76}\)
\(\frac{1}{77}< \frac{1}{76}\)
...................
\(\frac{1}{100}< \frac{1}{76}\)
\(\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{76}+\frac{1}{76}+...+\frac{1}{76}=25.\frac{1}{76}< 25.\frac{1}{75}=\frac{1}{3}\)(4)
Từ (3) và (4) ta có:
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{5}{6}\)(6)
Từ (5) và (6)
\(\Rightarrow\frac{7}{12}< \frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}< \frac{5}{6}\)
đpcm
Tham khảo nhé~
D
16 tháng 4 2015
b.Đặt A = \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+....+\frac{1}{100^2}\) < \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)= \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}\)= \(\frac{1}{4}-\frac{1}{100}=\frac{25}{100}-\frac{1}{100}=\frac{24}{100}\frac{1}{6}\)(2)
Từ (1) và (2) =>\(\frac{1}{6}\) < A < \(\frac{1}{4}\)
25 tháng 5 2015
Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}....\frac{100}{101}\)
Nhận xét: Nếu \(\frac{a}{b}
S
18 tháng 5 2017
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
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Câu hỏi của Tuấn Anh - Toán lớp 7 - Học toán với OnlineMath
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