設數列{an}的前n項和為Sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*).(1)...
問題詳情:設數列{an}的前n項和為Sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*).(1)求a2,a3的值;(2)求*:數列{Sn+2}是等比數列.【回答】解:(1)因為a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*),所以當n=1時,a1=2×1=2;當n=2時,a1+2a2=(a1+a2)+4,所以a2=4;當n=3時,a1+2a2+3...