\(B=4.3+4.3^2+...+4.3^{100}=4.\left(3+3^2+...+3^{100}\right)=4.\dfrac{3^{101}-3}{2}=2.3^{101}-6\)

\(a.\dfrac{15}{13}\times\dfrac{6}{17}+\dfrac{7}{15}\times\dfrac{15}{13}=\dfrac{15}{13}\times\left(\dfrac{6}{17}+\dfrac{7}{15}\right)=\dfrac{15}{13}\times\dfrac{6\times15+7\times17}{17\times15}=\dfrac{6\times15+7\times17}{13\times17}=\dfrac{209}{221}\)

\(b.\dfrac{1}{2}\times\dfrac{7}{8}\times\dfrac{6}{7}\times\dfrac{2}{3}=\dfrac{1\times7\times6\times2}{2\times3\times8\times7}=\dfrac{6\times7\times2}{6\times7\times8}=\dfrac{2}{8}=\dfrac{1}{4}\)