{an}等差数列,其公差为d 则
Sn＝a1＋(a1＋d)＋(a1＋2d)＋…＋[a1＋(n－1)d],
Sn＝an＋(an－d)＋(an－2d)＋…＋[an－(n－1)d],
两式相加得 2Sn＝n(a1＋an) ∴ Sn＝n(a1＋an) ／2