已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，且\(S_{n}+ \dfrac {1}{2}a_{n}=1(n∈N^{+}).\)
\((1)\)求数列\(\{a_{n}\}\)的通项公式；
\((2)\)设\(b_{n}=\log _{ \frac {1}{3}}(1-S_{n})(n∈N^{+})\)，求\( \dfrac {1}{b_{1}b_{2}}+ \dfrac {1}{b_{2}b_{3}}+…+ \dfrac {1}{b_{n}b_{n+1}}\)的值．