B=1x5+2x6+...+99x103
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\(1\cdot5+2\cdot6+\cdots+41\cdot45\)
\(=1\cdot\left(1+4\right)+2\cdot\left(2+4\right)+\cdots+41\cdot\left(41+4\right)\)
\(=\left(1^2+2^2+\cdots+41^2\right)+4\left(1+2+\cdots+41\right)\)
\(=\frac{41\left(41+1\right)\left(2\cdot41+1\right)}{6}+4\cdot\frac{41\cdot42}{2}\)
\(=\frac{41\cdot42\cdot83}{6}+2\cdot41\cdot42=41\cdot7\cdot83+2\cdot41\cdot42=27265\)
1x5+2x6+3x7+4x8+...+50x54
=(1+2+3+4+...50)x(5+6+7+78+...+54)
=1275x1269=1617975
\(A=\frac{2}{3\times7}+\frac{2}{7\times11}+\frac{2}{11\times15}+...+\frac{2}{99\times103}\)
\(2\times A=\frac{4}{3\times7}+\frac{4}{7\times11}+\frac{4}{11\times15}+...+\frac{4}{99\times103}\)
\(=\frac{7-3}{3\times7}+\frac{11-7}{7\times11}+\frac{15-11}{11\times15}+...+\frac{103-99}{99\times103}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{99}-\frac{1}{103}\)
\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow A=\frac{50}{309}\)