∵{an}为等差数列，其前n项之和为Sn，∴S2n-1=(2n−1)(a1+a2n−1)2=(2n−1)×2an2=（2n-1）•an，同理可得，S′2n-1=（2n-1）•bn，∴anbn=S2n−1S2n−1′，又SnS′n=7n+2n+3，∴S2n−1S′2n−1=7(2n−1)+2(2n−1)+3...