设数列An的前n项和为Sn,已知a(1)+2a(2)+3a(3)+…+na(n)=(n-1)Sn+2n(n为正整数）.求证数列Sn+2是等比数列

题目

设数列An的前n项和为Sn,已知a(1)+2a(2)+3a(3)+…+na(n)=(n-1)Sn+2n(n为正整数）.求证数列Sn+2是等比数列

答案

由原式得a1+2a2+3a3```+nan=nSn-Sn+2n a1+2a2+3a3```+nan+(n+1)a(n+1)=nS(n+1)+2n+2 ∴nSn-Sn+2n+(n+1)a(n+1)=nS(n+1)+2n+2 (注意a1+2a2+3a3```+nan=nSn-Sn+2na1+2a2+3a3```+nan+(n+1)a(n+1)=nS(n+1)+2n+2nSn-Sn+2n+...