数列{an}满足a1=1,an=2an-1-3n+6(n>=2,n∈N+)(1)设bn=an-3n,求证:数列{bn}是

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数列{an}满足a1=1,an=2an-1-3n+6(n>=2,n∈N+)(1)设bn=an-3n,求证:数列{bn}是等比数列

证：
n≥2时,
an=2a(n-1)-3n+6
an-3n=2a(n-1)-6(n-1)=2[a(n-1)-3(n-1)]
(an -3n)/[a(n-1)-3(n-1)]=2,为定值.
a1-3=1-3=-2,数列{an -3n}是以-2为首项,2为公比的等比数列.
bn=an-3n,数列{bn}是是以-2为首项,2为公比的等比数列.
再问：求{an}的前n项和Sn
再答： an-3n=-2×2^(n-1)=-2ⁿ an=3n-2ⁿ Sn=a1+a2+...+an =3(1+2+...+n)-(2+2^2+...+2ⁿ) =3n(n+1)/2 -2×(2ⁿ-1)/(2-1) =3n(n+1)/2 -2^(n+1) +2

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