数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝.

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数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝.
数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝.

数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝.
待定系数法
因为a（n+1）=2an-n^2+3n
设 a(n+1)+p(n+1)^2+q(n+1)=2(an+pn^2+qn)
展开整理得
a(n+1)=2an+pn^2+(q-2p)-(p+q)
与原式一一对应
所以p=-1 q=1
所以数列{an-n^2+n}为一个公比为2的等比数列且首相为1
所以an-n^2+n=2^（n-1)
an=2^（n-1)+n^2-n

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