已知数列{an}的通项公式是an=2n-1,bn=a1+a2+···+an(n属于n+) 求bn,设cn=【b（n+1）-bn】/3*n

因为an=2n-1,bn=a1+a2+···+an,所以bn=n^2,cn=(2n+1)/(3^n),所以Sn=3/(3^1)+5/(3^2)+:::+(2n+1)/(3^n),1/3Sn=3/(3^2)+::::+(2n+1)/(3^(n+1)),两式相减,得2/3Tn=1+2((1/3)^2+(1/3)^3+::::+(1/3)^n)-(2n+1)/(3^(n+1))=4/3-(1/3)^n-(2n+1)/(3^(n+1)),Tn=2-1/2×(1/3)^(n-1)-(2n+1)/(2×(3^n))cn=(2n+1)/(3^n)是哪来的呀！cn=b(n+1)-bn/3^n(n+1)是小标(b(n+1)-bn)/3^n=((n+1)^2-n^2)/3^n=(2n+1)/(3^n),不对么，还是我把你题目理解错了?我可以重做谢谢！谢谢！谢谢！这也算追问？呵呵，没事了吧