设数列{an}对所有正整数n都满足：a1+2a2+2^2a3+…+2^(n-1)an=8-5n,求数列{an}的前n项和Sn

由：a1+2a2+2^2a3+…+2^(n-1)an=8-5n--------------------------------①知：a1+2a2+2^2a3+…+2^(n-2)a(n-1)=8-5(n-1)-----------------------②①-②得：2^(n-1)an=-5 n≥2∴ an=-5*(1/2)^(n-1) n≥2a1=8-5=3∴S1=a...