(1)由数列的通项公式,能求出数列前三项.
(2)由由数列的通项公式,知Sn=(3+2×1+1)+(32+2×2+1)+(33+2×3+1)+…+(3n+2n+1),由此利用分组求和法和等比数列与等差数列的前n项和公式能求出结果.
【解析】
(1)∵数列的通项公式,
∴a1=3+2×1+1=6,
=32+2×2+1=14,
a3=33+2×3+1=34.
(2)Sn=(3+2×1+1)+(32+2×2+1)+(33+2×3+1)+…+(3n+2n+1)
=(3+32+33+…+3n)+2(1+2+3+…+n)+n
=+2×+n
=.