已知数列｛an｝满足a(n+1)=2an+n^2,a1=2bn=an+n^2+2n+3,(n∈N*)(1)求证{bn}为等比数列(2)求{an}通项公式

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已知数列｛an｝满足a(n+1)=2an+n^2,a1=2bn=an+n^2+2n+3,(n∈N*)(1)求证{bn}为等比数列(2)求{an}通项公式
已知数列｛an｝满足a(n+1)=2an+n^2,a1=2
bn=an+n^2+2n+3,(n∈N*)
(1)求证{bn}为等比数列(2)求{an}通项公式

已知数列｛an｝满足a(n+1)=2an+n^2,a1=2bn=an+n^2+2n+3,(n∈N*)(1)求证{bn}为等比数列(2)求{an}通项公式
(1)因为bn=an+n^2+2n+3,a(n+1)=2an+n^2,所以b(n+1)=a(n+1)+(n+1)^2+2(n+1)+3=2an+2n^2+4n+6
b(n+1)/bn=2
所以{bn}为等比数列.
(2)因为a1=2,所以b1=a1+1^2+2+3=8
所以bn=8*2^(n-1)=2^(n+2)
所以an=bn－2n－3＝2^(n+2)－2n－3

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