数列{an}满足a1=1.a(n+1)√{1/(an)2+4}=1 记Sn=a12+a22+.S（2n-1）-Sn≤m/

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数列{an}满足a1=1.a(n+1)√{1/(an)2+4}=1 记Sn=a12+a22+.S（2n-1）-Sn≤m/30恒成立求正整数m最小值
a12为a1的平方
Sn=a12+a22+，，，，，+an2
S（2n-1）-Sn≤m/30恒成立
我现在可以求出{1/an2}等差、an2=1/(4n-3)

感觉是“S(2n+1)-Sn≤m/30”吧?
∵数列{a[n]}满足a[n+1]√（1/a[n]^2+4)=1
∴1/a[n+1]^2-1/a[n]^2=4
∵a[1]=1
∴{1/a[n]^2}是首项为1/a[1]^2=1,公差为4的等差数列
即：1/a[n]^2=1+4(n-1)=4n-3
∴a[n]^2=1/(4n-3)
∵S[n]=a[1]^2+a[2]^2+……+a[n]^2
∴(S[2n+1]-S[n])-(S[2n+3]-S[n+1])
=(a[n+1]^2+a[n+2]^2+...+a[2n+1]^2)-(a[n+2]^2+a[n+3]^2+...+a[2n+1]^2+a[2n+2]^2+a[2n+3]^2)
=a[n+1]^2-a[2n+2]^2-a[2n+3]^2
=1/(4n+1)-1/(8n+5)-1/(8n+9)
∵1/(8n+2)>1/(8n+5),1/(8n+2)>1/(8n+9),1/(4n+1)=1/(8n+2)+1/(8n+2)
∴(S[2n+1]-S[n])-(S[2n+3]-S[n+1])>0
即：S[2n+1]-S[n]>S[2n+3]-S[n+1]
说明{S[2n+1]-S[n]}是一个递减数列
∴{S[2n+1]-S[n]}最大项为：S[3]-S[1]=a[2]^2+a[3]^3=1/5+1/9=14/45
∵S[2n+1]-S[n]≤m/30对n属于N*恒成立
∴S[2n+1]-S[n]≤S[3]-S[1]≤m/30
即：14/45≤m/30
解得：m≥28/3
∵n∈N*
∴m最小值是10
如果是S(2n-1)-Sn≤m/30的话
[S(2n-1)-Sn]-[S(2n+1)-S(n+1)]
=1/(4n+1)-1/(8n-3)-1/(8n+1)
＜0
是单调递增数列没有最大值

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