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6 tháng 3 2019

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)

\(=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)< \frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right)\)

\(=\frac{1}{4}\left(1-\frac{1}{n}\right)\)(đpcm)

6 tháng 3 2019

Ta có:\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)

\(=\frac{1}{4.4}+\frac{1}{4.9}+\frac{1}{4.16}+...+\frac{1}{4.n^2}\)

\(=\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)\)

\(Xét:\)

\(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};\frac{1}{n.n}< \frac{1}{\left(n-1\right).n}...\)

\(Suyra:\)

\(P=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

\(\Leftrightarrow P< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(\Leftrightarrow P< 1-\frac{1}{n}< 1\)

\(\Leftrightarrow\frac{1}{4}.P< 1.\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)< \frac{1}{4}\)

\(\Leftrightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\left(đpcm\right)\)

4 tháng 5 2016

Đặt A= \(\frac{1}{4^2}\) + \(\frac{1}{6^2}\) + \(\frac{1}{8^2}\) +...+ \(\frac{1}{\left(2n\right)^2}\)

A= \(\frac{1}{2^2.2^2}\) + \(\frac{1}{2^2.3^2}\) +...+ \(\frac{1}{2^2.n^2}\)

A= \(\frac{1}{2^2}\).( \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + ...+ \(\frac{1}{n^2}\))

A< \(\frac{1}{2^2}\) . ( \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) +...+ \(\frac{1}{\left(n-1\right)n}\)

A< \(\frac{1}{4}\) . ( 1-\(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) +...+ \(\frac{1}{n-1}\) - \(\frac{1}{n}\) )

A< \(\frac{1}{4}\) . (1-\(\frac{1}{n}\)) = \(\frac{1}{4}\) - \(\frac{1}{4n}\) <\(\frac{1}{4}\) => A <\(\frac{1}{4}\)

22 tháng 7 2016

Ta có : 

\(N=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

Ta thấy : \(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

.......

\(\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1\)

\(\Rightarrow\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)< 1.\frac{1}{2^2}\)

\(\Rightarrow N< \frac{1}{4}\)(ĐPCM)

Ủng hộ mk nha !!! ^_^

29 tháng 4 2017

Câu 1 :
 A = (2012+2) . [ ( 2012-2) : 3+1 ] : 2 = 2014 . 671 : 2 = 675697
 B = \(\frac{1}{2}\).  \(\frac{2}{3}\).  \(\frac{3}{4}\)+...+  \(\frac{2010}{2011}\).  \(\frac{2011}{2012}\)\(\frac{1.2.3.....2010.2011}{2.3.4.....2011.2012}\)=  \(\frac{1}{2012}\)
Câu 2 :
 a) \(2x.\left(3y-2\right)+\left(3y-2\right)=-55\)
=> \(\left(3y-2\right).\left(2x+1\right)=-55\)
=>  \(3y-2;2x+1\in\: UC\left(-55\right)\)
=>  \(3y-2;2x+1=\left\{1;-1;5;-5;11;-11;55;-55\right\}\)
- Vậy ta có bảng 

BẢNG TÌM x;y
\(2x+1\) 1-1 5-511-1155-55
\(x\) 0-1 2-35-627-28
\(3y-2\)-5555-1111-55-11
\(3y\)-5357-913-3713
\(y\)\(\frac{-53}{3}\)(loại)19(chọn)-3(chọn)\(\frac{13}{3}\)(loại)-1(chọn)\(\frac{7}{3}\)(loại)\(\frac{1}{3}\)(loại)1(chọn)


\(\Leftrightarrow\)Những cặp (x;y) tìm được là : 
(-1;19)  ;   (2;-3)   ;    (5;-1)    ;    (-28;1)
b) Ta đặt vế đó là A
Ta xét A :   \(\frac{1}{4^2}\)<  \(\frac{1}{2.4}\)
                  \(\frac{1}{6^2}\)<  \(\frac{1}{4.6}\)
                  \(\frac{1}{8^2}\)<  \(\frac{1}{6.8}\)
                          ...
                 \(\frac{1}{\left(2n\right)^2}\)<  \(\frac{1}{\left(2n-2\right).2n}\)

  \(\Leftrightarrow\)A < \(\frac{1}{2.4}\)+  \(\frac{1}{4.6}\)+...+  \(\frac{1}{\left(2n-2\right).2n}\)
  \(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{2}{2.4}\)+  \(\frac{2}{4.6}\)+...+  \(\frac{2}{\left(2n-2\right).2n}\))
  \(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{1}{2}\)-  \(\frac{1}{4}\)+  \(\frac{1}{4}\)-  \(\frac{1}{6}\)+...+  \(\frac{1}{2n-2}\)-  \(\frac{1}{2n}\))
  \(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{1}{2}\)-  \(\frac{1}{2n}\)) = \(\frac{1}{2}\).  \(\frac{1}{2}\)-  \(\frac{1}{2}\).  \(\frac{1}{2n}\)
  \(\Leftrightarrow\)A < \(\frac{1}{4}\)-  \(\frac{1}{4n}\)<  \(\frac{1}{4}\) ( Vì n \(\in\)N )
  \(\Leftrightarrow\)A <  \(\frac{1}{4}\)( đpcm ) .

29 tháng 4 2017

Bạn Phùng Quang Thịnh làm đúng hết rồi 

26 tháng 8 2016

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2n-2\right)2n}\)

\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}< \frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{\left(2n-2\right)2n}\)\(.\frac{1}{2}\)       Ta gọi là A

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2n-2\right)2n}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n}\right)=\frac{1}{2}.\frac{1}{2}-\frac{1}{2}.\frac{1}{2n}=\frac{1}{4}-\frac{1}{2n.2}\)

\(\Rightarrow M< \frac{1}{4}-\frac{1}{2n.2}< \frac{1}{4}\)

\(\Rightarrow M< \frac{1}{4}\left(Đpcm\right)\)

\(\)

 

 

25 tháng 9 2018

Bạn tham khảo cách làm ở đây: https://olm.vn/hoi-dap/question/528628.html