等差数列列｛an｝的前n项和为Sn,已知limSn/n^2= -(a1/9)

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等差数列列｛an｝的前n项和为Sn,已知limSn/n^2= -(a1/9)
等差数列列｛an｝的前n项和为Sn,已知limSn/n^2= -(a1/9)<0,问当n为何值时,Sn最大?为什么?

等差数列列｛an｝的前n项和为Sn,已知limSn/n^2= -(a1/9)
Sn=n*a1+(n-1)nd/2
Sn/n²=a1/n+(1 -1/n)d/2
∴Sn/n²=0+d/2=d/2=-(a1/9)＜0
∴d＜0,a1＞0,d=(-2/9)*a1
Sn=n*a1+(n-1)nd/2
=na1+(1-n)na1/9
=a1*[n+(n-n²)/9]
=a1*(10n-n²)/9
容易知道此函数看做二次函数时,开口向下,对称轴为n=5
∴n=5时,Sn取得最大值

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