\(B=\dfrac{1}{2\cdot15}+\dfrac{1}{15\cdot3}+...+\dfrac{6}{87\cdot90}\)
\(=\dfrac{6}{12\cdot15}+\dfrac{6}{15\cdot18}+...+\dfrac{6}{87\cdot90}\)
\(=2\left(\dfrac{3}{12\cdot15}+\dfrac{3}{15\cdot18}+...+\dfrac{3}{87\cdot90}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{18}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{90}\right)=2\cdot\left(\dfrac{15}{180}-\dfrac{2}{180}\right)=2\cdot\dfrac{13}{180}=\dfrac{13}{90}\)
\(\dfrac{1}{2.15}\)+\(\dfrac{1}{15.3}\)+\(\dfrac{1}{3.21}\)+...........+\(\dfrac{6}{87.90}\)
= 6.(\(\dfrac{1}{12.15}\)+\(\dfrac{1}{15.18}\)+\(\dfrac{1}{18.21}\)+.............+\(\dfrac{1}{87.90}\))
= \(\dfrac{6}{3}\).(\(\dfrac{1}{12}\)−\(\dfrac{1}{15}\)+\(\dfrac{1}{15}\)−\(\dfrac{1}{18}\)+\(\dfrac{1}{18}\)−\(\dfrac{1}{21}\)+..........+\(\dfrac{1}{87}\)−\(\dfrac{1}{90}\))
= \(\dfrac{6}{3}\).(\(\dfrac{1}{12}\)−\(\dfrac{1}{90}\))
= \(\dfrac{6}{3}\).\(\dfrac{13}{180}\)
= \(\dfrac{13}{90}\)