\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2n-1.2n+1}=\frac{49}{99}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{49}{99}\)
\(1-\frac{1}{2n+1}=\frac{49}{99}\)
\(\frac{1}{2n-1}=1-\frac{49}{99}\)
\(\frac{1}{2n-1}=\frac{50}{99}\Rightarrow99=50\left(2n-1\right)\)
=>99=100n-1
=>-1-99=-100n
=>-100=-100n
=>n=1
1/1.3+1/3.5+1/5.7+.....+1/(2n-1).(2n+1)=49/99
1/2.(2/1.3+2/3.5+2/5.7+.....+2/(2n-1).(2n+1))=49/99
1/2(1/1-1/3+1/3-1/5+1/5-1/7+......+1/2n-1-1/2n+1=49/99
1/2.(1/1-2n+1)=49/99
1/2n+1=49/99:1/2
1/2n+1=98/99
\(\Rightarrow\)1.98/2n+1.98=98/99
\(\Rightarrow\)2n+1.98=99
2n+1=99:98
2n+1=99/98
n+1=99/98:2
n+1=99/192
n=99/192-1
n=-97/192
Vế trái là
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)=\(2\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left[2n-1\right]\left[2n+1\right]}\right)\)=\(2<1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}>=2\left(1-\frac{1}{2n+1}\right)\)
\(1-\frac{1}{2n+1}=\frac{49}{99}:2\)
\(\frac{1}{2n+1}=1-\frac{49}{198}\)
Còn lại tự làm nốt