已知等差数列an满足a3=7,a5+a7=26,an的前n项和为sn.1.求an和sn

1：an=kn+b
7=3k+b
26=5k+b+7k+b
k=2 b=1
an=2n+1
sn=2（1+2+3+.+n）+n=n(n+1)+n=n^2+2n
2:an的平方-1=(2n+1-1)(2n+1+1)=4n(n+1)
bn=1/4n(n+1)=1/4(1/n-1/n+1)
Tn=1/4(1-1/n+1)
=1/(4n+4)