CMR : \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)\(< 2\)
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13 tháng 8 2019
Ta có:
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{25}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(=\frac{1}{25}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{25}+\frac{1}{6}-\frac{1}{101}>\frac{1}{6}+\frac{1}{25}-\frac{1}{100}=\frac{1}{6}+\frac{3}{100}>\frac{1}{6}\left(1\right)\)
\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\left(2\right)\)
Từ (1) và (2) suy ra:\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(đpcm\right)\)
MN
13 tháng 8 2019
đạt 1/52+.........+1/1002=S
1/52>1/5*6
.....................
1/1002>1/100*101
=>S>1/5*6+.............+1/100*101=1/5-1/6+....+1/100-1/101=1/5-1/101=96/505>96/576=1/6
vậ S>1/6
1/52<1/4*5
.....................
1/1002<1/99*100
=>S<1/4*5+................+1/99*100=1/4-1/5+.....+1/99-1/100=1/4-1/100=6/25<6/24=1/4
Vậy 1/6<S<1/4
23 tháng 10 2016
a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(=\frac{1}{5}+\frac{2}{7}\)
\(=\frac{7}{35}+\frac{10}{35}\)
\(=\frac{17}{35}\)
Vậy \(A=\frac{17}{35}\)
b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)
\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)
\(=5.\frac{50}{671}\)
\(=\frac{250}{671}\)
Vậy \(B=\frac{250}{671}\)
NV
16 tháng 1 2019
Ta có:
\(\frac{1}{5^2}>\frac{1}{5.6}\)
\(\frac{1}{6^2}>\frac{1}{6.7}\)
.......
\(\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\) \(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\) = \(\frac{1}{5}-\frac{1}{101}>\frac{1}{5}-\frac{1}{30}=\frac{1}{6}\) \(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{6}\) (1)
Tương tự ta có:
\(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
......
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\) \(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4}\) (2)
Từ (1) và (2)
\(\Rightarrow\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4}\) (đpcm)
_Chúc_bạn_học_tốt_
2
1
S
7 tháng 5 2017
Đặt A la tên của biểu thức trên
\(A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+...+2017}\)
\(=\frac{1}{2\left(3+1\right):2}+\frac{1}{3\left(5+1\right):2}+\frac{1}{4\left(7+1\right):2}+...+\frac{1}{1009\left(2017+1\right):2}\)
\(=\frac{2}{2.4}+\frac{2}{3.6}+\frac{2}{4.8}+....+\frac{2}{1009.2018}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{1009.1009}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}\right)\)
Ta có: \(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...........
\(\frac{1}{1009^2}< \frac{1}{1008.1009}\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{1008.1009}\right)\)
\(A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1008}-\frac{1}{1009}\right)\)
\(A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{1009}\right)=\frac{1}{4}+\frac{1}{2}-\frac{1}{1009}=\frac{3}{4}-\frac{1}{1009}< \frac{3}{4}\)
Vậy ...
7 tháng 5 2017
Đặt tổng đã cho là A
\(\frac{1}{1+3}=\frac{1}{\left(3+1\right)x2:2}=\frac{1}{2x4:2}=\frac{1}{2x4}x2=\frac{2}{2x4}\)=\(\frac{1}{2x2}\)
\(\frac{1}{1+3+5}=\frac{1}{\left(1+5\right)x3:2}=\frac{1}{3x6}x2=\frac{2}{3x6}\)=\(\frac{1}{3x3}\)
\(\frac{1}{1+3+5+....+2017}=\frac{1}{\left(1+2017\right)x1009:2}=\frac{1}{1009x2018}x2=\frac{2}{1009x2018}\)=\(\frac{1}{1009x1009}\)
Các mẫu là bạn áp dụng tính tổng đó nha ( mk làm tắt)
A=\(\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{1009x1009}\)<\(\frac{1}{2x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{1008x1009}=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1008}-\frac{1}{1009}\)=\(\frac{1}{4}+\frac{1}{2}-\frac{1}{1009}< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)
vậy A<3/4( Mk có làm tắt nên chỗ nào ko hiểu thì nhắn tin nha
Ta thấy : \(0< \frac{1}{5}< 1\)
\(1< \frac{1}{5}>\frac{1}{6}>.....>\frac{1}{17}>0\)
Viết lại :
\(2>\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{17}>1>0\)
Mik ko biết có đúng ko