\(B=\frac{1}{1x2x5}+\frac{1}{2x3x4}+\frac{1}{3x4x5}+...+\frac{1}{98x99x5}+\frac{1}{99x100x5}\)
\(5B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{98x99}+\frac{1}{99x100}\)
\(5B=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{99-98}{98x99}+\frac{100-99}{99x100}\)
\(5B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(5B=1-\frac{1}{100}=\frac{99}{100}\Rightarrow B=\frac{99}{500}\)