(1)a(n+2)-2a(n+1)=a(n+1)-2an
即b(n+1)=bn=...=b1=a2-2a1=1
所以{bn}为常数列
(2)a(n+1)-2an=1
所以a(n+1)+1=2(an+1)
即{an+1}为等比数列,公比为2,而a1+1=3
an+1=3*2^(n-1)
an=3*2^(n-1)-1
(3)Sn=3(1-2^n)/(1-2)-n
=3(2^n-1)-n
数列{an}满足a1=2,a2=5,a（n+2）-3a(n+1)+2an=0.
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