chứng minh rằng
\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{7^2}\)+...+\(\frac{1}{2017^2}\).,<\(\frac{1}{4}\)
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29 tháng 5 2017
bài này dài lắm
\(A=\frac{\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}}{\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}}\)
\(A=\frac{\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)}{\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)}\)
\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)}\)
\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)}\)
\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}\)
\(A=\frac{\left(\frac{1}{100}\right)}{\left(\frac{1}{25}\right)}=\frac{1}{4}\)
\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{125}{42}}{\frac{2000}{43}-\frac{250}{252}-\frac{2000}{257}}\)
\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{6000}{2016}}{\frac{2000}{43}-\frac{2000}{2016}-\frac{2000}{257}}\)
\(B=\frac{16.\left(\frac{1}{9}-\frac{1}{127}+\frac{1}{2017}\right)}{5.\left(\frac{1}{2017}+\frac{1}{9}-\frac{1}{127}\right)}-\frac{6000.\left(\frac{1}{43}-\frac{1}{257}-\frac{1}{2016}\right)}{2000.\left(\frac{1}{43}-\frac{1}{2016}-\frac{1}{257}\right)}\)
\(B=\frac{16}{5}-3=\frac{1}{5}\)
Đặt \(C=\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}\)
\(C=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2005^2}+\frac{1}{2006^2}+\frac{1}{2007^2}\)
\(C< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2004.2005}+\frac{1}{2005.2006}+\frac{1}{2006.2007}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2006}+\frac{1}{2006}-\frac{1}{2007}\)
\(=\frac{1}{4}-\frac{1}{2017}\left(đpcm\right)\)
\(C>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2005.2006}+\frac{1}{2006.2007}+\frac{1}{2007.2008}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2007}-\frac{1}{2008}\)
\(=\frac{1}{5}-\frac{1}{2008}\left(đpcm\right)\)
Vậy \(A>\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}>B\)
D
15 tháng 1 2020
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
16 tháng 12 2019
\(Ta\) \(có : \)
\(1 / 5^2 + 1 /6^2 + ... + 1 /100^2 < 1 /4.5\)\(+ 1 / 5 .6 + ... + 1 / 99 .100\)
\(Mà ta có:\)\(1 / 4 .5 + 1 / 5 .6 + ... + 1 / 99 .100\)
\(\Rightarrow\)\(1 / 4 - 1 / 5 + 1 / 5 - 1 / 6 + ... +\)\(1 / 99 - 1 / 100\)
\(\Rightarrow\)\(1 / 4 - 1 / 100\) \(< 1 / 4\)
\(Nên 1 / 5^2 + 1 /6^2 + ...+ 1 / 100^2 < 1 / 4\)
Tương tự chứng minh tiếp nhé 😘😘
27 tháng 11 2016
C/m<1/4
t\(n^2>n\left(n-1\right)=>\frac{1}{n^2}<\frac{1}{n\left(n-1\right)}\)
\(\frac{1}{5^2}<\frac{1}{4.5};\frac{1}{6^2}<\frac{1}{5.6};\frac{1}{100^2}<\frac{1}{99.100}\)
\(\frac{1}{4.5}+..+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}.ok\)
CM>1/6
\(n^2\frac{1}{n\left(n+1\right)}\)
\(\frac{1}{5.6}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{5}>\frac{1}{6}\)OK
27 tháng 11 2016
doan cuoi
\(\frac{1}{5}-\frac{1}{101}=\frac{96}{5.101}>\frac{96}{5.102}=\frac{1.}{6}.\frac{96}{85}>\frac{1}{6}ok\)
21 tháng 4 2016
ta có :\(\frac{1}{5^2}<\frac{1}{4.5}\)
\(\frac{1}{6^2}<\frac{1}{5.6}\)
\(\frac{1}{7^2}<\frac{1}{6.7}\)
.....
\(\frac{1}{100^2}<\frac{1}{99.100}\)
\(\Rightarrow A<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\) (1)
Ta có : \(\frac{1}{5.6}<\frac{1}{5^2}\)'
\(\frac{1}{6.7}<\frac{1}{6^2}\)
....\(\frac{1}{100.101}<\frac{1}{100^2}\)
\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\) <A
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{101}\) <A
\(\frac{1}{5}-\frac{1}{101}\) <A
mà \(\frac{96}{5.101}=\frac{96}{505}>\frac{96}{576}\)
hay \(A>\frac{1}{6}\) (2)
từ (1); và (2) suy ra \(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+..+\frac{1}{100^2}<\frac{1}{4}\) (đpcm)
đây là cách dễ hiểu nhất nhé
Ta có :
\(\frac{1}{5^2}< \frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}=\frac{1}{5}-\frac{1}{6}\)
\(\frac{1}{7^2}< \frac{1}{6.7}=\frac{1}{6}-\frac{1}{7}\)
\(....\)
\(\frac{1}{2017^2}< \frac{1}{2016.2017}=\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+....+\frac{1}{2017^2}< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=\frac{1}{4}-\frac{1}{2017}< \frac{1}{4}\) (đpcm)
Ta có: 1/52 + 1/62 + ... + 1/20172 < 1/4.5 + 1/5.6 + ... + 1/2016.2017
Mà: 1/4.5 + 1/5.6 + ... + 1/2016.2017 = 1/4 - 1/2017
=> 1/52 + 1/62 + ... + 1/20172 < 1/4