求数列-1,4,7,10...(-1)^n(3n-2)的前n项和.急!!!!!!

来源：百度知道 编辑：UC知道 时间：2024/06/23 22:40:37
RT.过程!!!谢谢!!!!

当n为奇数时，
Sn=-1+4-7+10-13+16+...+(3n-5)-(3n-2)
=-(1+7+13+...+3n-2)+(4+10+16+...+3n-5)
=-(1+3n-2)/2*(n+1)/2+(4+3n-5)/2*(n-1)/2
=(3n-1)/4*[(n-1)-(n+1)]
=(3n-1)/4*(-2)
=-(3n-1)/2
当n为偶数时，
Sn=-1+4-7+10-13+16+...-(3n-5)+(3n-2)
=-(1+7+13+...+3n-5)+(4+10+16+...+3n-2)
=-(1+3n-5)/2*n/2+(4+3n-2)/2*n/2
=n/4*[(3n+2)-(3n-4)]
=n/4*6
=3n/2

分低！