作业帮数列：A(n+1)^2+An^2+16=8[A(n+1)+An]+2A(n+1)An,则An=?
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/02 21:00:56
![数列：A(n+1)^2+An^2+16=8[A(n+1)+An]+2A(n+1)An,则An=?](/uploads/image/z/11178358-70-8.jpg?t=%E6%95%B0%E5%88%97%EF%BC%9AA%28n%2B1%29%5E2%2BAn%5E2%2B16%3D8%5BA%28n%2B1%29%2BAn%5D%2B2A%28n%2B1%29An%2C%E5%88%99An%3D%3F)
数列：A(n+1)^2+An^2+16=8[A(n+1)+An]+2A(n+1)An,则An=?
数列：A(n+1)^2+An^2+16=8[A(n+1)+An]+2A(n+1)An,则An=?
数列：A(n+1)^2+An^2+16=8[A(n+1)+An]+2A(n+1)An,则An=?
A(n+1)^2+An^2+16=8[A(n+1)+An]+2A(n+1)*An
A(n+1)^2+A(n+2)^2+16=8[A(n+1)+A(n+2)]+2A(n+1)*A(n+2)
A(n+2)^2-An^2=8*(A(n+2)-An)+2A(n+1)*(A(n+2)-An)
(A(n+2)-An)*(A(n+2)+An)=8*(A(n+2)-An)+2A(n+1)*(A(n+2)-An)
(A(n+2)-An)*(A(n+2)+An-2A(n+1)-8)=0
A(n+2)-A(n+1)=A(n+1)-An+8
这题少条件,如果知道A1,就能得出An了……