作业帮等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__
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![等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__](/uploads/image/z/4045574-38-4.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E5%89%8Dn%E9%A1%B9%E5%92%8CSn+%E5%B7%B2%E7%9F%A5lim+%5BSn%2F%28n%26%23178%3B%2B1%EF%BC%89%5D%3D-a1%2F8%EF%BC%88a1%3E0%29+%E5%88%99Sn%E8%BE%BE%E5%88%B0%E6%9C%80%E5%A4%A7%E5%80%BC%E6%97%B6%E7%9A%84n%3D__)
等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__
等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__
等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__
设an=a1+(n-1)d
有 Sn=na1+n(n-1)d/2
limSn/(n^2+1)
=lim[na1+n(n-1)d/2]/(n^2+1)
=lim[a1/n+d/2-d/(2n)]/(1+/n^20)
=[0+d/2-0]/(1+0)
=d/2
所以d/2=-a1/8,d=-a1/4
an=a1+(n-1)d=a1-(n-1)a1/4
显然将所有的正数或非负数相加可得最大值,如果再加负数,则会减小,所以
an=a1-(n-1)a1/4≥0
因为a1>0,所以1-（n-1)/4≥0
n≤5,
n=5时,a5=0
所以Sn达到最大值时的n=4,或n=5.
等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__
等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__
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