给你推一下吧.设此数列为 a,a+d,a+2d,a+3d ,...,a+(n-1)d,则前n项的和为
Sn=a+(a+d)+(a+2d)+(a+3d)+...+[a+(n-1)d]
=n*a+d[1+2+3+...+(n-1)]
=n*a+n*(n-1)*d/2