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](/uploads/image/z/2815989-69-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3a%28n%2B1%29%3D2an%2Bn%5E2%2Ca1%3D2bn%3Dan%2Bn%5E2%2B2n%2B3%2C%28n%E2%88%88N%2A%29%281%29%E6%B1%82%E8%AF%81%7Bbn%7D%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%282%29%E6%B1%82%7Ban%7D%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
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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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