答案:
①依题意,根据等差数列通项公式,an+an+1=3n-2,
当n>1时,an+an-1=3n-5,an+1-an-1=3,
即a2n-1(n∈N*)和a2n(n∈N*)都是公差为3的等差数列.
因为a1=0,a2=1,
所以a2n-1=3(n-1),a2n=3n-2,
即an=,k∈N*.
②Sn=2-1×a1+2-2×a2+2-3×a3++2-n+1×an-1+2-n×an2Sn=a1+2-1×a2+2-2×a3++2-n+2×an-1+2-n+1×an,
两式相加得3Sn=2-1(a2+a1)+2-2(a3+a2)++2-n+2(an-1+an-2)+2-n+1(an+an-1)+2-n×an6Sn
=(a2+a1)+2-1(a3+a2)++2-n+3(an-1+an-2)+2-n+2(an+an-1)+2-n+1×an
两式相减得:3Sn=1+2-1×3+2-2×3++2-n+2×3-2-n+1×(3n-5)+2-n×an3Sn
=1+3×(1-2-n+2)-2-n+1×(3n-5)+2-n×an=4-2-n×(6n+2-an)
Sn=,
所以Sn=,k∈N*.