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
问题描述:
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)
答
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...