（1）设等差数列的首项是a1,公差为d
则 a1+2d=7 ①
2a1+10d=26 ②
①*5-②
解得 a1=3,∴ d=2
∴ an=a1+(n-1)d=2n+1
∴ Sn=(a1+an)*n/2=(2n+4)*n/2=n(n+2)
（2）bn=an平方-1分之1
=1/[(2n+1)²-1]
=1/[2n(2n+2)]
=(1/4)*1/[n(n+1)]
=(1/4)*[1/n-1/(n+1)]
∴ 数列（bn）的前n项和Tn为：
Tn=(1/4)*[1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)]
=(1/4)*[1-1/(n+1)]
=(1/4)*n/(n+1)