已知数列{an}中,a1=1,Sn=3a(n+1) (1)求｛an}通项an (2)求a2+a4+a6+.+a2n 注：

1) Sn=3a(n+1)
=> S(n-1)=3an
=> Sn-S(n-1)=3a(n+1)-3an=an,
=> a(n+1)=4an/3
=> {an}为首项a1=1,公比q=4/3的等比数列
=> an=a1×qⁿ/q=(4/3)^(n-1)
2) a2+a4+a6+……+a2n
=(4/3)+(4/3)³+(4/3)^5+……+(4/3)^(2n-1)
=(4/3)×[(4/3)º+(4/3)²+(4/3)⁴+……+(4/3)^(2n-2)]
=(4/3)×{1-[(4/3)²]ⁿ}/[1-(4/3)²]
=(-12/7)×{1-[(4/3)²]ⁿ}