Question

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及Sn; (2)令bn=1/

Answer 1

1)a5+a7=2a6=26得a6=13

d=(a6-a3)/(6-3)=(13-7)/3=2

a1=a3-2d=3

Sn=3n+(1/2)n(n-1)·2=n^2+2n

2)an=2n+1,然后写出bn的式子,利用拆项求和,得Tn=n/[4(n+1)]