已知数列\(\{a_{n}\}\)满足\(\dfrac{n}{a_{1}+a_{2}+\cdots+a_{n}}=\dfrac{1}{4n+1}(n\in N^{*})\)，若数列\(\{b_{n}\}\)满足\(b_{n}=\dfrac{a_{n}+3}{4}\)，则\(\dfrac{1}{b_{1}b_{2}}+\dfrac{1}{b_{2}b_{3}}+…+\dfrac{1}{b_{2020}b_{2021}}=(\quad)\)