(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, ∴ 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= 故存在常数k=及等差数列an=n-使其满足题意…………………12分 |