记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn求证：数列（Sn/n)为等比数列求_数学解答

记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn
求证：数列（Sn/n)为等比数列
求数列（an）的通项公式

人气：410 ℃　时间：2020-09-03 08:09:44

解答

证明,因为A(n+1) = (n+2)/n * Sn所以Sn = n*A(n+1) / (n+2)S(n-1) = (n-1)*An / (n+1)所以An = Sn - S(n-1) = n/(n+2) *A(n+1) - (n-1)/(n+1) * An所以2n/(n+1) * An = n/(n+2) * A(n+1)即A(n+1)/An = (2n+4)/(n+1)...