已知数列{an}是等差数列,它的前n项和为Sn.a1+a2+a3=4,a3+a4+a5=10.求SNrtttttttttttttt_数学解答

已知数列{an}是等差数列,它的前n项和为Sn.a1+a2+a3=4,a3+a4+a5=10.求SN
rtttttttttttttt

人气：237 ℃　时间：2019-10-10 05:12:55

解答

a1+a2+a3=4
a1+a1+d+a1+2d=4
3a1+3d=4
a3+a4+a5=10
a1+2d+a1+3d+a1+4d=10
3a1+9d=10
与3a1+3d=4连立
解得d=1 a1=1/3
所以
Sn=na1+d*n(n-1)/2
=n/3+n(n-1)/2
=(3n^2-n)/6