(1)由数列{an}、{bn}满足a1=1,a2=3,,bn=an+1-an,知{bn}是以2为首项,以2为公比的等比数列,由此求出.从而利用累加法能求出an=2n-1.
(2)由an=2n-1,bn=2n-1,知cn=bn•log2(an+1)==n•2n-1,由此利用错位相减法能求出Sn=c1+c2+…+cn.
【解析】
(1)∵数列{an}、{bn}满足a1=1,a2=3,,bn=an+1-an,
∴b1=a2-a1=3-1=2,
∴{bn}是以2为首项,以2为公比的等比数列,
∴.
∴a2-a1=2,
a3-a2=22,
…
an-an-1=2n-1,
∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)
=1+2+22+…+2n-1
=
=2n-1.
(2)∵an=2n-1,bn=2n-1,
∴cn=bn•log2(an+1)==n•2n-1,
∴Sn=c1+c2+…+cn=1×2+2×2+3×22+…+n•2n-1,
∴2Sn=1×2+2×22+3×23+…+n•2n,
∴-Sn=1+2+22+23+…+2n-1-n•2n
=-n•2n
=2n-1-n•2n,
∴-2n+1.