已知数列\(\{a _{n} \}\)满足\( \dfrac {1}{2a_{1}+3}+ \dfrac {2}{2a_{2}+3}+ \dfrac {3}{2a_{3}+3}+…+ \dfrac {n}{2a_{n}+3}= \dfrac {n}{2}(n∈N^{*})\)．
\((\)Ⅰ\()\)求数列\(\{a _{n} \}\)的通项公式；
\((\)Ⅱ\()\)设\(b_{n}= \dfrac {1}{a_{n}a_{n+1}}\)，求数列\(\{b _{n} \}\)的前\(2020\)项和\(T _{2020}\)．