a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{667}{668}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{667}{668}\)
\(1-\frac{1}{x+1}=\frac{667}{668}\)
\(\frac{1}{x+1}=1-\frac{667}{668}\)
\(\frac{1}{x+1}=\frac{1}{668}\)
\(\Rightarrow x+1=668\)
x = 667
a) 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/x.(x+1) = 667/668
=>1/1-1/2+1/2-1/3+1/3-1/4+.......+1/x-1/x+1=667/668
=>1/1-1/x+1=667/668
=>1/x+1=1/1-667/668
=>1/x+1=1/668
=>x=667
b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{667}{2002}\)
\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(1-\frac{1}{x+3}=\frac{2001}{2002}\)
\(\frac{1}{x+3}=\frac{1}{2002}\)
=> x + 3 = 2002
x = 1999
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{667}{668}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{667}{668}\)
\(=1-\frac{1}{x+1}=\frac{667}{668}\)
\(=1-\frac{667}{668}=\frac{1}{668}\)
\(\frac{1}{x+1}=\frac{1}{668}\Rightarrow x+1=668\Rightarrow x=667\)
b) \(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{667}{2002}\)
\(\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{66}{2002}\)
\(\frac{1}{3}\cdot\left(1-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(1-\frac{1}{x+3}=\frac{2001}{2002}\Rightarrow\frac{1}{x+3}=\frac{1}{2002}\Rightarrow x+3=2002\Rightarrow x=1999\)
