Đặt tổng trên = A
\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\)
\(A.2=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\)
\(A.2-A=\left(\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\right)-\left(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\right)\)
\(A=\frac{4}{3}-\frac{2}{192}\)
\(A=\frac{127}{96}\)
\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\)
\(2A=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\)
\(2A-A=\frac{4}{3}-\frac{2}{192}\)
\(A=\frac{4}{3}-\frac{2}{192}=\frac{127}{96}\)