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+,,,,,+an2S(2n-1)-Sn≤m/30恒成立我现在可以求出{1/an2}等差、an2=1/(4n-3)
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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+,,,,,+an2S(2n-1)-Sn≤m/30恒成立我现在可以求出{1/an2}等差、an2=1/(4n-3)](/uploads/image/z/8822702-38-2.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1.a%28n%2B1%29%E2%88%9A%7B1%2F%28an%292%2B4%7D%3D1+%E8%AE%B0Sn%3Da12%2Ba22%2B.S%EF%BC%882n-1%EF%BC%89-Sn%E2%89%A4m%2F30%E6%81%92%E6%88%90%E7%AB%8B+%E6%B1%82%E6%AD%A3%E6%95%B4%E6%95%B0m%E6%9C%80%E5%B0%8F%E5%80%BCa12%E4%B8%BAa1%E7%9A%84%E5%B9%B3%E6%96%B9Sn%3Da12%2Ba22%2B%EF%BC%8C%EF%BC%8C%EF%BC%8C%EF%BC%8C%EF%BC%8C%2Ban2S%EF%BC%882n-1%EF%BC%89-Sn%E2%89%A4m%2F30%E6%81%92%E6%88%90%E7%AB%8B%E6%88%91%E7%8E%B0%E5%9C%A8%E5%8F%AF%E4%BB%A5%E6%B1%82%E5%87%BA%7B1%2Fan2%7D%E7%AD%89%E5%B7%AE%E3%80%81an2%3D1%2F%284n-3%29)
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+,,,,,+an2S(2n-1)-Sn≤m/30恒成立我现在可以求出{1/an2}等差、an2=1/(4n-3)
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)
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+,,,,,+an2S(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