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

\(\frac{1}{4^2}<\frac{1}{3.4}\)

...

\(\frac{1}{100^2}<\frac{1}{99.100}\)

===>\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<\frac{50}{100}=\frac{1}{2}\)

\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2016^2}\)

\(=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{1008^2}\right)< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1007.1008}\right)\)

                                                                         \(< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1007}-\frac{1}{1008}\right)\)

                                                                           \(< \frac{1}{4}.\left(2-\frac{1}{1008}\right)< \frac{1}{4}.2=\frac{1}{2}\)

=> đpcm

có: 1/3^2<1/2.3; 1/4^2<1/3.4:...: 1/100^2<1/99.100

Mà: 1/1.2+1/2.3+...+1/99.100=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

=> 1/3^2+1/4^2+...+1/100^2<99/100<1

=> đpcm

UNDERSTAND ???

đặt A= biểu thức trên

tao có 

A<1/2.3+1/3.4+...+1/99.100

A<1/2-1/3+1/3-1/4+...+1/99-1/100

A<1/2-1/100<1/2

SUY RA A<1/2(DPCM)

a) Để â nhận giá trị nguyên

\(\Rightarrow8n-9⋮2n+5\)

\(\Rightarrow8n+20-29⋮2n+5\)

\(\Rightarrow4.\left(2n+5\right)-29⋮2n+5\)

mà \(4.\left(2n+5\right)⋮2n+5\)

\(\Rightarrow-29⋮2n+5\)

\(\Rightarrow2n+5\inƯ\left(-29\right)\)

tự làm nốt nhé, tick nha

khó quá!!!Bó tay...Sorry

`1/4+1/16+1/36+...+1/196`

`= 1/(2^2)+1/(4^2)+1/(6^2)+....+1/(4^2)`

`= 1/(2^2)*( 1/ + 1/( 2^2 ) + 1/(3^2)+.....+1/(7^2))`

Ta có : `1/(2^2)<1/(1*2)=1-1/2`

`1/(3^2)<1/(2*3)=1/2-1/3`

`.....`

`1/(7^2)<1/(6*7)=1/6-1/7`

Do `1/( 2^2 ) + 1/(3^2)+.....+1/(7^2)<1-1/2+1/2-1/3+.....+1/6-1/7=1-1/7<1`

`=> 1/ + 1/( 2^2 ) + 1/(3^2)+.....+1/(7^2)<2`

`=>  1/(2^2)*( 1/ + 1/( 2^2 ) + 1/(3^2)+.....+1/(7^2))<1/2`

`=>1/4+1/16+1/36+...+1/196<1/2`

Vậy `1/4+1/16+1/36+....+1/196<1/2` 

tự làm cũng phải ghi à bạn 

Hình như sai đề thì phải chứ mk làm ko đc !!!

  A < 1/(1.2) + 1/(2.3) + 1/(3.4) + ...+ 1/(99.100) 
<=> A< 1- 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + .. + 1/99 - 1/100 
<=> A < 1 - 1/100 < 1 (đpcm) 

So với  thì đây