\(A=\frac{2}{3^2}+\frac{2}{5^2}+.......+\frac{2}{2007^2}\)

\(A=2.\left(\frac{1}{3.3}+\frac{1}{5.5}+......+\frac{1}{2007.2007}\right)\)

\(A< 2.\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2006.2007}\right)\)

\(A< 2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2006}-\frac{1}{2007}\right)\)

\(A< 2.\left(\frac{1}{2}-\frac{1}{2007}\right)\)

\(A< 2.\frac{2005}{4014}\)

\(A< \frac{2005}{2007}\)

Ta thấy

2/(3x3) < 2/(2x4)  = 1/2 – 1/4

2/(5x5) < 2/(4x6) = 1/4 – 1/6

2/(7x7) < 2/(6x8) = 1/6 – 1/8

………

2/(2007x2007) < 2/(2006x2008) = 1/2006 – 1/12008

Nên:

A = 2/3^2 +2/5^2+2/7^2 +.....+2/2007^2 < 2/(2x4) + 2/(4x6) + …. + 2/(2006x2008) =

1/2 – 1/4 + 1/4 – 1/6 + 1/6 – 1/8 + … + 1/2006 – 1/2008 =          

1/2 – 1/2008 = 1003/2008

Vậy: .....

A < 2005/2007

\(A< \frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2007.2009}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}=\frac{1}{3}-\frac{1}{2009}=\frac{2006}{6027}< \frac{2006}{4016}=\frac{1003}{2008}\)Vây:.......

a) \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}<\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{2006\cdot2007}\)

=>              \(<\frac{1}{4}-\frac{1}{2007}<\frac{1}{4}\)

\(vậy:\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{2007^2}<\frac{1}{4}\)

b) \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{2007\cdot2008}\)

=>    \(>\frac{1}{5}-\frac{1}{2008}>\frac{1}{5}\)

\(vậy:\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5}\)

cảm ơn bạn nha

Ta thấy: 3²>3²-1=(3-1).(3+1)=2.4

              5²>5²-1=(5-1).(5+1)=4.6

              7²>7²-1=(7-1).(7+1)=6.8

              …………………………

              2007²>2007²-1=(2007-1).(2007+1)=2006.2008

=>          \(\frac{2}{3^2}

bạn Lê Quốc Vượng cũng chơi bang bang hả có những tank gì rồi .Tớ có tank Triệu Vân, joker,tiên

cá, doraemon, quan công, nhện, pea,pega

bạn có thích nhạc nơi này có anh 

<3 <3 <3

\(A=2\cdot\left(\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2017^2}\right)< 2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}\right)\)

Đặt \(M=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)

\(\Rightarrow M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)

\(\Rightarrow M=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}< \frac{1}{1009}+\frac{1}{1009}+...+\frac{1}{1009}\)(1008 số hạng )

hay\(M< \frac{1008}{1009}\Rightarrow A< 2\cdot\frac{1008}{1009}=\frac{504}{1009}\left(ĐPCM\right)\)

a)S=1+(-1/7)^1+(-1/7)^2+...+(-1/7)^2007

=>7S=7+(-1/7)^1+(1/7)^2+...+(-1/7)^2006

=>(7-1)S=6-(1/7)^2007

=>S=1-(-1/7^2007/6)