求等差数列5,7,9...2n+7 的各项之和

本题先得求出共有多少项，即：n=[（2n+7）-5]/2+1=n+2 其中除数2为公差
然后就用求和公式：Sn=[5+(2n+7)]*(n+2)/2=n^2+8n+12

sn=n(5+2n+7)/2=n(6+n)

Sn=5+7+9+----+2n+7
=(5+2n+7)[(2n+7-5)/2+1]/2
=(n+2)(n+6)
=n^2+8n+12

首项为n=1，即2n-7=9，Sn=(9+2n-7)n/2+7+5=(2n'2+(a-7)n)/2+5+7