(1)设{an}的公差为d,由题意得a1=1,d=2, 所以an=2n-1,Sn=na1+ d=n2. (2)b1=a1=1,bn+1=bn+an=bn+2n-1, 所以b2=b1+1,b3=b2+3=b1+1+3, bn=b1+1+3+…+(2n-3)=1+(n-1)2=n2-2n+2(n≥2). 又n=1时n2-2n+2=1=b1, 所以数列{bn}的通项公式为bn=n2-2n+2. (3)cn= = = - , Tn=c1+c2+…+cn=( - )+( - )+…+( - )=1- = . |