设公差为d的等差数列{an}的前k1,k2,…,kn项的和,依次为Sk1,Sk2,…,Skn.当k1,k2,…,kn是公比为q(|q|≠1)的等比数列时,则称Sk1+Sk2+…+Skn为等差数列的等比累进和,并记为SSn.若令m=k1(1-qn)1-q,m′=k1(1+qn)1+q,则有如下的定理SSn=ma1+12m(m′-1)d.证∵公差为d的等差数列{an}的 Let the sum of the first k1, k2, ..., kn terms of the arithmetic difference sequence {an} of tolerance d be Sk1,Sk2,...,Skn. When k1,k2,...,kn is the common ratio q(|q When the geometric series of ≠ 1) is equal, then Sk1+Sk2+...+Skn is called the geometric progression sum of the arithmetic progression and denoted as SSn. If let m=k1(1-qn)1-q, m′= K1(1+qn)1+q, there is the following theorem SSn=ma1+12m(m′-1)d.