(1)在等差数列{an}中,Sn,S2n-Sn,S3n-S2n,…成等差数列, ∴Sn+(S3n-S2n)=2(S2n-Sn) ∴S3n=3 S2n-3 Sn=60…………………………………………………………………4分 (2)SpSq= pq(a1+ap)(a1+aq) = pq[a +a1(ap+aq)+apaq] = pq(a +2a1am+apaq)< ( )2[a +2a1am+( )2] = m2(a +2a1am+a )=[ m(a1+am)]2 =S ………………………………………………………………………8分 (3)设an=pn+q(p,q为常数),则ka -1=kp2n2+2kpqn+kq2-1 Sn+1= p(n+1)2+ (n+1) S2n=2pn2+(p+2q)n ∴S2n-Sn+1= pn2+ n-(p+q), 依题意有kp2n2+2kpqn+kq2-1= pn2+ n-(p+q)对一切正整数n成立, ∴![](http://img.shitiku.com.cn/uploads/allimg/20191011/20191011130603-34397.gif) 由①得,p=0或kp= ; 若p=0代入②有q=0,而p=q=0不满足③, ∴p≠0 由kp= 代入②, ∴3q= ,q=- 代入③得,
-1=-(p- ),将kp= 代入得,∴P= , 解得q=- ,k=![](http://img.shitiku.com.cn/uploads/allimg/20191011/20191011130600-41287.gif) 故存在常数k= 及等差数列an= n- 使其满足题意…………………12分 |