Bài 9:
\(\frac12\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{n\left(n+2\right)}\right)=\frac{2000}{2001}\)
=>\(\frac12\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{\left(n+1\right)^2-1}\right)=\frac{2000}{2001}\)
=>\(\frac12\cdot\frac{2^2}{2^2-1}\cdot\frac{3^2}{3^2-1}\cdot\ldots\cdot\frac{\left(n+1\right)^2}{\left(n+1\right)^2-1}=\frac{2000}{2001}\)
=>\(\frac12\cdot\frac{2\cdot3\cdot\ldots\cdot\left(n+1\right)}{1\cdot2\cdot\ldots\cdot n}\cdot\frac{2\cdot3\cdot\ldots\cdot\left(n+1\right)}{3\cdot4\cdot\ldots\cdot\left(n+2\right)}=\frac{2000}{2001}\)
=>\(\frac12\cdot\frac{n+1}{1}\cdot\frac{2}{n+2}=\frac{2000}{2001}\)
=>\(\frac{n+1}{n+2}=\frac{2000}{2001}\)
=>\(1-\frac{1}{n+2}=1-\frac{1}{2001}\)
=>n+2=2001
=>n=1999
