数列{an},a1=1,an=2-2Sn,求an,若bn=n*an,求{bn}的前n项和Tn

问题描述:

数列{an},a1=1,an=2-2Sn,求an,若bn=n*an,求{bn}的前n项和Tn

因为an=2-2Sn……(1)所以a(n-1)=2-2S(n-1)……(2)(1)-(2)得:an- a(n-1)= -2(Sn-S(n-1))即an- a(n-1)= -2an推出于an=(1/3)a(n-1)所以是以1为首项1/3为公比的等比数列所以an=(1/3)^(n-1)则bn=n*(1...