数列{an}满足a1=1,an+1*根号（1/an^2+4)=1,S2n+1-Sn

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∵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
【问题补充时,楼主将第一个“+”号写成“=”,应该是“笔”误,这里改过来.】
∴(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
∴m的最小值是28/3

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