\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)
=> x + 1 = 2003
=> x = 2003 - 1
=> x = 2002
13+16+110+...+1x(x+1):2=2001200313+16+110+...+1�(�+1):2=20012003
26+212+220+...+2x(x+1)=2001200326+212+220+...+2�(�+1)=20012003
2.(12.3+13.4+14.5+...+1x(x+1))=200120032.(12.3+13.4+14.5+...+1�(�+1))=20012003
12−13+13−14+14−15+...+1x−1x+1=20012003:212−13+13−14+14−15+...+1�−1�+1=20012003:2
12−1x+1=2001400612−1�+1=20014006
=> 1x+1=12−20014006=120031�+1=12−20014006=12003
=> x + 1 = 2003
=> x = 2003 - 1
=> x = 2002
