设数列An满足a1=0且1/(1-An+1)-1/(1-An)=1求An的通项公式?
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![设数列An满足a1=0且1/(1-An+1)-1/(1-An)=1求An的通项公式?](/uploads/image/z/5000164-52-4.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97An%E6%BB%A1%E8%B6%B3a1%3D0%E4%B8%941%2F%281-An%2B1%29-1%2F%281-An%29%3D1%E6%B1%82An%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%3F)
设数列An满足a1=0且1/(1-An+1)-1/(1-An)=1求An的通项公式?
设数列An满足a1=0且1/(1-An+1)-1/(1-An)=1求An的通项公式?
设数列An满足a1=0且1/(1-An+1)-1/(1-An)=1求An的通项公式?
∵ 1/(1-An+1) - 1/(1-An) = 1
∴数列{1/(1-An)} 是以1/(1-A1) = 1 为首项,1为公差的等差数列.
∴1/(1-An) = 1 + (n-1)×1 = n
∴An = 1 - 1/n = (n-1)/n