11.已知数列{an}\{a_{n}\}{an}的前nnn项和Sn=3+2nS_{n}=3+2^{n}Sn=3+2n,则数列{an}\{a_{n}\}{an}的通项公式为___an={5,(n=1)2n−1,(n⩾2){a_n}=\left\{\begin{array}{l}5,(n=1)\\ {2^{n-1}},(n\geqslant 2)\end{array}\right.an={5,(n=1)2n−1,(n⩾2)___.