已知数列{an}的前n项和Sn=2n^2+2n,数列{bn}的前n项和Tn=2-bn

(1)n=1时,S1=1-a1 所以a1=1/2an=Sn-S(n-1)=1-an-(1-a(n-1)=a(n-1)-an所以：an=1/2a(n-1),{an}是等比数列an=(1/2)^n(2)Tn =2*1/2+3*(1/2)^2+……+(n+1)*(1/2)^n1/2Tn=2*(1/2)^2+……+n*(1/2)^n+(n+1)*(1/2)^(n+1)【这...