已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，且\(S_{n}= \dfrac {n^{2}}{2}+ \dfrac {3n}{2}\)．
\((1)\)求数列\(\{a_{n}\}\)的通项公式；
\((2)\)若数列\(\{b_{n}\}\)满足\(b_{n}=a_{n+2}-a_{n}+ \dfrac {1}{a_{n+2}\cdot a_{n}}\)，且数列\(\{b_{n}\}\)的前\(n\)项和为\(T_{n}\)，求证：\(T_{n} < 2n+ \dfrac {5}{12}\)．