Đặt A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}\)

A=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}\)

A = \(1-\frac{1}{n+2}\)

A= \(\frac{n+1}{n+2}\)=> Để A<2003/2004 thì \(\left(n+1\right).2004< \left(n+2\right).2003\)

\(\Leftrightarrow2004n+2004< 2003n+4006\)

\(\Leftrightarrow n< 2002\)

1/1-1/3+1/3-1/5+1/5-1/7+....+1/n-1/(n+2)

=1-1/(n+2)=(n+1)/(n+2)

Suy ra n =2001

<=>2-2/3+2/3-2/5........+2n-2n+2<2015/2016

<=>2-2n+2<2015/2016

=>n+2=1/2016

=>n=2014

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{n\left(n+2\right)}\)<\(\frac{2015}{2016}\)

VT=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{5}-\frac{1}{n+2}\)=\(1-\frac{1}{n+2}\)

Ta có:\(1-\frac{1}{n+2}=\frac{2015}{2016}\Rightarrow\)\(\frac{1}{n+2}=1-\frac{2015}{2016}\)

\(\Rightarrow\)\(\frac{1}{n+2}=\frac{1}{2016}=n+2=2016\)

\(\Rightarrow\)\(n=2014\)

Vậy\(n=2014\)

\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\)

\(=1-\frac{1}{x+2}\frac{1}{2016}\Rightarrow x+2

A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)

\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)

\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)

X=16

17 - 1= 16

= > x = 16

 tk mình nha

17 - 1= 16

= > x = 16

 tk mình nha

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{n\left(n+2\right)}< \frac{2003}{2004}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{n}+\frac{1}{n+2}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{n+2}\right)\)

\(=\frac{1}{2}\left(\frac{n+2}{n+2}-\frac{1}{n+2}\right)\)

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

\(=\frac{n+1}{2\left(n+2\right)}< \frac{2003}{2004}\)

\(\Leftrightarrow\hept{\begin{cases}n+1< 2003\\2\left(n+2\right)< 2004\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}n< 2002\\\left(n+2\right)< 1002\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}n< 2002\\n< 1000\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}n+1=2002\\2\left(n+2\right)=1000\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}n=2001\\n=498\end{cases}}\)

Lời giải :

program hotrotinhoc ;

var T : real ;

i,n : integer ;

begin

write('n='); readln(n);

S:=0;

for i:= 1 to n do

if i mod 2=1 then

S:=S+2/(i*(i+2));

write('Tong la :',S);

readln

end.

uses crt;
var a:array[1..10000000] of longint;
i,n,d:longint;
s:real;
begin
clrscr;
write('Nhap n la ');readln(n);
for i:=1 to n do begin
if i mod 2<>0 then
s:=s+(2/(i*(i+2))); end;
write('Tong la ',s:0:2);
readln
end.