(1)由a1=1,=2,=3可得a2=2a1,a3=3a2,a4=2a3,可求a3+a4
(2)由已知可得a2k+1=3a2k=3(2a2k-1)=6a2k-1,则数列的奇数项是以1为首项,以6为公比的等比数列;由a2k=2a2k-1即偶数项都是前一项的2倍,从而对n分类讨论:分n=2k时,当n=2k-1两种情况,利用等比数列的求和公式分别求解
【解析】
(1)∵a1=1,=2,=3
∴a2=2a1=2,a3=3a2=6,a4=2a3=12
∴a3+a4=18
(2)∵a1=1,=2,=3
∴a2k+1=3a2k=3(2a2k-1)=6a2k-1
∴数列的奇数项是以1为首项,以6为公比的等比数列
∵a2k=2a2k-1即偶数项都是前一项的2倍
当n=2k时,Sn=a1+a2+a3+…+an
=1+2×1+6+2×6+62+2×62+…++2×
=3(1+6+…+)
==
当n=2k-1时,Sn=a1+a2+a3+…+an
=1+2×1+6+2×6+…+-2×=
故答案为:18;