An为等差数列,且满足A(n+1)=An²-nAn+1,n=1,2,3……求An的通项公式
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An为等差数列,且满足A(n+1)=An²-nAn+1,n=1,2,3……求An的通项公式
An为等差数列,且满足A(n+1)=An²-nAn+1,n=1,2,3……求An的通项公式
An为等差数列,且满足A(n+1)=An²-nAn+1,n=1,2,3……求An的通项公式
a(n+1)=an^2-nan+1
an=a(n-1)^2-(n-1)a(n-1)+1
两式相减
d=(2a(n-1)+d)*d-nd-a(n-1)
d=2a1d+d^2-2d-a1
d=2a2d+d^2-3d-a2
两式相减
2d^2-d-d=0
得d1=0,d2=1
当d=0时 解得a1=1 an=1
当d=1时 解得a1=0或2
当a1=0时 an=n-1 代入式An²-nAn+1=2-n≠n (舍弃)
当a1=2时 an=n+1 代入式An²-nAn+1=n+2=a(n+1)符合题意
所以 an=1 或an=n+1``````````
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