\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+...+\frac{2}{x\left(x+3\right)}=\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}\)
\(=\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Từ đó ta có:
\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(x+3=308\)
x = 305