B = \(\frac{6}{1.3}+\frac{6}{3.5}+...+\frac{6}{97.99}=3.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
=\(3.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)=3.\left(1-\frac{1}{99}\right)=3.\frac{98}{99}=\frac{98}{33}\)
\(B=\frac{6}{1.3}+\frac{6}{3.5}+...+\frac{6}{97.99}\)
\(=3\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(=3\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(1-\frac{1}{99}\right)\)
\(=\frac{98}{33}\)
2/3B=2/1.3+2/3.5+...+2/97.99
2/3B=1-1/3+1/3-1/5+.....+1/97-1/99
2/3B=1-1/99
2/3B=98/99
B=98/99:2/3
B=294/198
