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
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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