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数列：A(n+1)^2+An^2+16=8[A(n+1)+An]+2A(n+1)An,则An=?
题目详情
数列：A(n+1)^2+An^2+16=8[A(n+1)+An]+2A(n+1)An,则An=?
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答案和解析
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……
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……
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