设数列的{an}的前项n和为Sn，若对于任意的正整数n都有Sn=2an－3n

解:a1=S1=2a1-3,得a1=3Sn=2an-3nS(n-1)=2a(n-1)-3(n-1)Sn-S(n-1)=2an-2a(n-1)-3=an所以an+3=2(a(n-1)+3)数列{an+3}是以a1+3=3+3=6为首项,以2为公比的等比数列an+3=6*2^(n-1),an=3*2^n-3bn=an+3=3*2^n{bn}是以b1=6为首项,2为公比的等比数列.设数列{nan}的前n项和为Tn则Tn=3(1*2+2*2^2+3*2^3+...+n*2^n)-3n(n+1)/22Tn=3( 1*2^2+2*2^3+...+(n-1)*2^n+n*2^(n+1)-3n(n+1)上式减去下式得:-Tn=3(2+2^2+2^3+...+2^n)-3n*2^(n+1)+3n(n+1)/2-Tn=3(2*(2^n-1))-3n*2^(n+1)+3n(n+1)/2所以Tn=3(n-1)2^(n+1)-3n(n+1)/2+6 收起