已知等差数列{an}公差为d,满足a1+a2+a3=15,且a1+1,a3+1,a7+1构成等你数列的连续三项（1）求数列{an}的通项公式（2）若等差数列{an}递增,其前n项和为Tn,求证1/3=
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题目中的等你数列也许是等比数列是吗?
(1)a1+a2+a3=3a2=15,a2=5,设公差为d ,a1=5-d,a3=5+d,a7=5+5d
(a3+1)^2=(a1+1)(a7+1) 所以(6+d)^2=(6-d)(6+5d),解得 d=2 或d=0
当d=2时,an=2n+1,当d=0时,an=5
(2)an=2n+1,Tn=(3+2n+1)*n/2=(n+2)*n
1/Tn=1/(n+2)*n=1/2[1/n-1/(n+2)],设求证式的中间一式为M
M=1/2[1-1/3
+1/2-1/4
+1/3-1/5
+1/4-1/6
+.+1/(n-1)-1/(n+1)
+1/n-1/(n+2)]=1/2[1+1/2-1/(n+1)-1/(n+2)]=3/4-1/2[1/(n+1)+1/(n+2)],很明显,M