记等差数列\(\{a _{n} \}\)的前\(n\)项和为\(S _{n}\)．
\((1)\)求证：数列\(\{ \dfrac {S_{n}}{n}\}\)是等差数列；
\((2)\)若\(a_{1}=1,\{ \sqrt {S_{n}}\}\)是公差为\(1\)的等差数列，求使\( \dfrac {S_{k+1}\cdot S_{k+2}}{S_{k}^{2}}\)为整数的正整数\(k\)的取值集合；
\((3)\)记\(b_{n}=t^{a_{n}} (t\)为大于\(0\)的常数\()\)，求证：\( \dfrac {b_{1}+b_{2}+……+b_{n}}{n}\leqslant \dfrac {b_{1}+b_{2}}{2}\)．