设数列{an}的前n项和为Sn.若对任意的正整数n,总存在正整数m,使得Sn=am,则称{an}是"H数列". (1)若数列{an}的前n项和为Sn=2n(n∈N*),证明:{an}是"H数列". (2)设{an}是等差数列,其首项a1=1,公差d<0,若{an}是"H数列",求d的值; (3)证明:对任意的等差数列{an},总存在两个"H数列" {bn}和{cn},使得an=bn+cn(n∈N*)成立.
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