设数列﹛an﹜的前n项和为Sn,已知a1=8,an+1=Sn+3^(n+1)+5,n∈N.设bn=an-23^n,证

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设数列﹛an﹜的前n项和为Sn,已知a1=8,an+1=Sn+3^(n+1)+5,n∈N*.设bn=an-2*3^n,证明﹛bn﹜是等比数列

a(n+1)=Sn+3^(n+1)+5
an=S(n-1)+3^n+5
a(n+1)-an=[Sn-S(n-1)]+3^(n+1)-3^n=an+2*3^n
a(n+1)-2*3^(n+1)=2an-4*3^n=2[an-2*3^n]
an-2*3^n=[a1-2*3^1]2^(n-1)=[a1-6]2^(n-1)=2^n
an=2*3^n+2^n
bn=an-2*3^n=bn=2*3^n+2^n-2*3^n=2^n
bn为首项为2公比为2的等比数列.

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