Question

\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...........+\(\dfrac{1}{x\left(x+1\right):2}\)=\(\dfrac{2001}{2003}\)

Answer 1

=>\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2001}{2003}\)

=>\(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2001}{4006}\)

=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)

=>\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)

=>1/(x+1)=1/2-2001/4006=1/2003

=>x+1=2003

=>x=2002