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anan+1-2an+1+1=0怎么变形得1/(1-an+1)-1/(1-an)=1
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anan+1-2an+1+1=0怎么变形得1/(1-an+1)-1/(1-an)=1
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不加下标读起来好痛苦啊,试着做了下,应该这样没错.有问题在问我,希望可以采纳,真的好辛苦啊.注意：[ ]表示下标
An*A[n+1]-2A[n+1]+1=0
An*A[n+1]-A[n+1]-A[n+1]+1=0
(An-1)*A[n+1]-(A[n+1]-1)=0 除以(An-1)×(A[n+1]-1)
A(n+1)/(A[n+1]-1) - 1/(An-1) = 0 带入A[n+1]= A[n+1]-1+1
（A[n+1]-1+1）/(A[n+1]-1) - 1/(An-1) = 0
1+ 1/(A[n+1]-1) - 1/(An-1) = 0 1移到右边
1/(A[n+1]-1) - 1/(An-1) = -1 同时乘以-1
1/(1-A[n+1]) - 1/(1-An) = 1
An*A[n+1]-2A[n+1]+1=0
An*A[n+1]-A[n+1]-A[n+1]+1=0
(An-1)*A[n+1]-(A[n+1]-1)=0 除以(An-1)×(A[n+1]-1)
A(n+1)/(A[n+1]-1) - 1/(An-1) = 0 带入A[n+1]= A[n+1]-1+1
（A[n+1]-1+1）/(A[n+1]-1) - 1/(An-1) = 0
1+ 1/(A[n+1]-1) - 1/(An-1) = 0 1移到右边
1/(A[n+1]-1) - 1/(An-1) = -1 同时乘以-1
1/(1-A[n+1]) - 1/(1-An) = 1
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