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

cuijing2022-10-04 11:39:541条回答

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天羽翼共回答了18个问题 | 采纳率94.4%: 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; 1年前

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a(n)=a+(n-1)d,[a(n)]^2=[a+(n-1)d]^2=[nd+a-d]^2,
s(n)=na+n(n-1)d/2=(d/2)n^2+n(a-d/2),
s(n)/[a(n)]^2=[(d/2)n^2+n(a-d/2)]/[nd+a-d]^2=[d/2+(a-d/2)/n]/[d+(a-d)/n]^2,
lim(n->无穷大){s(n)/[a(n)]^2}=lim(n->无穷大）{[d/2+(a-d/2)/n]/[d+(a-d)/n]^2}=(d/2)/d^2=1/(2d)
d=2
lim(n->无穷大){s(n)/[a(n)]^2}=0.25

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