已知等差数列｛an｝的通向公式为an=3n-5,求其前n项和公式及S20,
问题描述：
已知等差数列｛an｝的通向公式为an=3n-5,求其前n项和公式及S20,
答
an=3n-5
a1=-2
a(n+1)-an=3(n+1)-5-[3n-5]=3
{an}是首项为a1=-2,公差为d=3的等差数列
Sn=na1+n(n-1)d/2或用公式Sn=n(a1+an)/2
=-2n+3n(n-1)/2 =n(-2+3n-5)/2
=3n^2/2-7n/2 =n(3n-7)/2
S20=3×20^2/2-7×20/2 S20 =20×(3×20-7)/2
=600-70 =10×53
=530 =530