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

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