已知等差数列｛an}的前n项之和为Sn,a1=1+根号2.S3=9+3根号2

(1) ∵2a2=a1+a3 ∴S3=3a2=9+3√2 a2=3+√2∴d=a2-a1=2 an=1+√2+(n-1)×2=2n+√2-1(2) ∵ Sn=na1+n(n-1)=n(n+√2) ∴bn=n+√2设任意三项p,q,r 成等比则(p+√2)(r+√2) = (q+√2)(q+√2) pr + (p+r)√2 = q^2+2q√2...