设数列{an}的前n项和为Sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*).(1)...
10-20
问题详情:设数列{an}的前n项和为Sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*).(1)求a2,a3的值;(2)求*:数列{Sn+2}是等比数列.【回答】解:(1)因为a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*),所以当n=1时,a1=2×1=2;当n=2时,a1+2a2=(a1+a2)+4,所以a2=4;当n=3时,a1+2a2+3...