数列{an}的前n项和为Sn,a1=1,an+1（n+1是下标）=2Sn+1,求数列{nan}的前n项和为Tn

a(n+1)=2Sn+1,an=2S(n-1)+1,相减为:a(n+1)-an=2an,a(n+1)=3an,q=3an=a1*q^(n-1)=3^(n-1)Tn=3^0+2*3^1+3*3^2+.+n*3^(n-1)3*Tn=3^1+2*3^2+3*3^3+.+n*3^n-2*Tn=Tn-3*Tn=[3^0+2*3^1+3*3^2+.+n*3^(n-1)]-[3^1+2*3^2+3*3^3...