证：
(1)
设公差为d.
Sn/n=S1/1 +(n-1)d=a1+(n-1)d
Sn=na1+n(n-1)d
S(n+1)=(n+1)a1+n(n+1)d
a(n+1)=S(n+1)-Sn=(n+1)a1+n(n+1)d-na1-n(n-1)d=a1+2nd
an=a1+2(n-1)d
a(n+1)-an=a1+2nd-a1-2(n-1)d=2d,为定值.
数列{an}是以2d为公差的等差数列.
(2)
S1=2 S2=6
d=S2/2 -S1/1=3 -2=1
Sn/n=S1/1 +(n-1)d=S1+(n-1)d=2+n-1=n+1
Sn=n(n+1)/2
1/Sn=2/[n(n+1)]=2[1/n-1/(n+1)]
前n项和Tn=2[1/1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)