记数列{an}的前n项和为Sn,若不等式an2+Sn2n2≥ma12对任意等差数列{an}及任意正整数n都成立,则实数m的最

2025-02-25 05:21:57
推荐回答(1个)
回答1:

an2+
Sn2
n2
=an2+
1
n2
[na1+
1
2
n(n-1)d]2
=an2+[a1+
1
2
(n-1)d]2
1
2
(n-1)d=t,
an2+
Sn2
n2
=(a1+2t)2+(a1+t)2
=2a12+6ta1+5t2
=5(t-
3a1
5
2+2a12-
9a12
5

当t=
3a1
5
时,取到最小值
1
2
(n-1)d=
3a1
5
,即n=
6a1
5d
+1,
∵不等式an2+
Sn2
n2
≥ma1
2对任意等差数列{an}及任意正整数n都成立,
∴m≤
1
5

∴实数m的最大值为
1
5

故选:D.