Question

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x\left(x-1\right)}=\frac{2007}{2009}\). Tim x

Answer 1

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x-1\right)}=\frac{2007}{2009}\)

\(2\cdot\left(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}\right)=\frac{2007}{2009}\)

\(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}=\frac{2007}{2009}:2\)

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)x}=\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{x-1}-\frac{1}{x}=\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{x}=\frac{2007}{4018}\)

\(\frac{1}{x}=\frac{1}{2}-\frac{2007}{4018}\)

\(\frac{1}{x}=\frac{1}{2009}\)

=> x = 2009