数列{an}的前n项和记为Sn,a1=t,点（Sn,a（n+1））（n+1为底数）在直线y=2x+1上,n∈N+

1) (y1-y2)/(x1-x2)=2
a(n+1)-an = 2an
a(n+1)/an = 3
a2 = 3t =2a1+1 = 2t+1
t = 1
2)an = 3^(n-1)
a(n+1) = 3^n
bn = n
Tn = 1/2 +1/6+.
= 裂项法...能具体点吗？我觉得已经够详细了裂项法..1/1 * 1/2 = 1 -1/2.1/2 * 1/3 = 1/2 - 1/31/3 * 1/4 = 1/3 - 1/4...以此类推，当它们相加，1/2,1/3....全部抵消，只剩首尾即：T2013 = 1 - 1/2013 = 2012/2013