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12 tháng 8 2016
Ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};\frac{1}{5^2}< \frac{1}{4.5};....;\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{49}{100}< \frac{1}{2}\)
Vậy \(C=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
NA
0
PL
3
28 tháng 6 2016
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(< 1-\frac{1}{100}< 1\)
=> đpcm
S
0
20 tháng 6 2017
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{10^2}\)
\(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\)
\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(D< 1-\dfrac{1}{10}\Leftrightarrow D< 1\left(đpcm\right)\)
PH
2
T
8 tháng 5 2018
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
\(\dfrac{1}{7^2}< \dfrac{1}{6.7}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
A<\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
A<\(\left[\left(\dfrac{1}{1}-\dfrac{1}{8}\right)+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(-\dfrac{1}{8}+\dfrac{1}{8}\right)\right]\)A<\(\left[\left(\dfrac{8}{8}-\dfrac{1}{8}\right)+0+0+0+0+0+0+0\right]\)
A<\(\dfrac{7}{8}< 1\)
Vậy ta có đpcm.
T
8 tháng 5 2018
Sorry nha, chỗ phân tích ra thành tống đại số phải như này :
A<\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
Rồi làm tương tự.
Mình cho bạn công thức nè :\(\dfrac{1}{n\left(n+1\right)}< \dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
4 tháng 5 2018
\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
Nên B<\(\dfrac{1}{4}\)
B=\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\)
Nên B>\(\dfrac{1}{6}\)
Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2013^2}\)
\(A=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+...+\frac{1}{2013\cdot2013}\)
Ta có : \(\frac{1}{5\cdot5}< \frac{1}{4\cdot5}\)
\(\frac{1}{6\cdot6}< \frac{1}{5\cdot6}\)
\(\frac{1}{7\cdot7}< \frac{1}{6\cdot7}\)
...
\(\frac{1}{2013\cdot2013}< \frac{1}{2012\cdot2013}\)
=> \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+..+\frac{1}{2013^2}< \frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{2012\cdot2013}\)
=> \(A< \frac{1}{4}-\frac{1}{2013}\)
=> \(A< \frac{2009}{8052}\)
Lại có \(\frac{2009}{8052}< \frac{1}{4}\)
Theo tính chất bắc cầu => \(A< \frac{1}{4}\)( đpcm )
Sai thì mong bạn bỏ qua