等差数列{an}的前n项和为Sn,且S4=40,Sn=210,Sn-4=130,则n=__S4=a1+a2+a3+a4=40 Sn-Sn-4=an+an-1+an-2+an-3=80∴4(a1+an)=120.∴a1+an=30.∴Sn=((a1+an)×n )/2 =(30×n )/2 =210.∴n=14.问:Sn-Sn-4=an+an-1+an-2+an-3=80 这一步怎么
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![等差数列{an}的前n项和为Sn,且S4=40,Sn=210,Sn-4=130,则n=__S4=a1+a2+a3+a4=40 Sn-Sn-4=an+an-1+an-2+an-3=80∴4(a1+an)=120.∴a1+an=30.∴Sn=((a1+an)×n )/2 =(30×n )/2 =210.∴n=14.问:Sn-Sn-4=an+an-1+an-2+an-3=80 这一步怎么](/uploads/image/z/1132000-16-0.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94S4%3D40%2CSn%3D210%2CSn-4%3D130%2C%E5%88%99n%3D__S4%3Da1%2Ba2%2Ba3%2Ba4%3D40+Sn-Sn-4%3Dan%2Ban-1%2Ban-2%2Ban-3%3D80%E2%88%B44%EF%BC%88a1%2Ban%EF%BC%89%3D120%EF%BC%8E%E2%88%B4a1%2Ban%3D30%EF%BC%8E%E2%88%B4Sn%3D%28%28a1%2Ban%29%C3%97n+%29%2F2+%3D%2830%C3%97n+%29%2F2+%3D210%EF%BC%8E%E2%88%B4n%3D14%EF%BC%8E%E9%97%AE%EF%BC%9ASn-Sn-4%3Dan%2Ban-1%2Ban-2%2Ban-3%3D80+%E8%BF%99%E4%B8%80%E6%AD%A5%E6%80%8E%E4%B9%88)
等差数列{an}的前n项和为Sn,且S4=40,Sn=210,Sn-4=130,则n=__S4=a1+a2+a3+a4=40 Sn-Sn-4=an+an-1+an-2+an-3=80∴4(a1+an)=120.∴a1+an=30.∴Sn=((a1+an)×n )/2 =(30×n )/2 =210.∴n=14.问:Sn-Sn-4=an+an-1+an-2+an-3=80 这一步怎么
等差数列{an}的前n项和为Sn,且S4=40,Sn=210,Sn-4=130,则n=__
S4=a1+a2+a3+a4=40
Sn-Sn-4=an+an-1+an-2+an-3=80
∴4(a1+an)=120.
∴a1+an=30.
∴Sn=((a1+an)×n )/2
=(30×n )/2
=210.
∴n=14.
问:Sn-Sn-4=an+an-1+an-2+an-3=80 这一步怎么得来的 thank you啦
等差数列{an}的前n项和为Sn,且S4=40,Sn=210,Sn-4=130,则n=__S4=a1+a2+a3+a4=40 Sn-Sn-4=an+an-1+an-2+an-3=80∴4(a1+an)=120.∴a1+an=30.∴Sn=((a1+an)×n )/2 =(30×n )/2 =210.∴n=14.问:Sn-Sn-4=an+an-1+an-2+an-3=80 这一步怎么
因为Sn=a1+a2+……+an-4+an-3+an-2+an-1+an
Sn-4=a1+a2+……+an-4
所以Sn-Sn-4=(a1+a2+……+an-4+an-3+an-2+an-1+an)-(a1+a2+……+an-4)=an+an-1+an-2+an-3
又Sn=210,Sn-4=130
所以Sn-Sn-4=210-130=80
所以Sn-Sn-4=an+an-1+an-2+an-3=80
S(n-4)=a1+a2+a3+……+a(n-4)
S4=a1+a2+a3+a4
Sn=a1+a2+a3+……+an
Sn-S(n-4)=an+a(n-1)+a(n-2)+a(n-3)=90
额,这个,Sn-Sn-4本来就是an+an-1+an-2+an-3,Sn是等差数列求和啊
数列前N项和就是这样啊