数列{an}中a1=2,对n∈N+,a(n+1)=(an/2)++(1/an),若an≤a恒成立,则a的取值范围
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![数列{an}中a1=2,对n∈N+,a(n+1)=(an/2)++(1/an),若an≤a恒成立,则a的取值范围](/uploads/image/z/12479659-43-9.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%ADa1%3D2%2C%E5%AF%B9n%E2%88%88N%2B%2Ca%28n%2B1%29%3D%28an%2F2%29%2B%2B%281%2Fan%29%2C%E8%8B%A5an%E2%89%A4a%E6%81%92%E6%88%90%E7%AB%8B%2C%E5%88%99a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
数列{an}中a1=2,对n∈N+,a(n+1)=(an/2)++(1/an),若an≤a恒成立,则a的取值范围
数列{an}中a1=2,对n∈N+,a(n+1)=(an/2)++(1/an),若an≤a恒成立,则a的取值范围
数列{an}中a1=2,对n∈N+,a(n+1)=(an/2)++(1/an),若an≤a恒成立,则a的取值范围
a(n+1)=(an/2)+(1/an)
2a(n+1)an=(an)^2+2
2[a(n+1)-√2]an=(an)^2-2√2an+2=(an-√2)^2
2[a(n+1)+√2]an=(an)^2+2√2an+2=(an+√2)^2
两式相除:
[a(n+1)-√2]/[a(n+1)+√2]=(an-√2)^2/(an+√2)^2
设bn=(an-√2)/(an+√2),b1=(a1-√2)/(a1+√2)=(1-√2)/(1+√2)=2√2-3
b(n+1)=(bn)^2
={[b(n-1)]^2}^2=[b(n-1)]^4
={[b(n-2)]^2}^4=[b(n-1)]^(2^3)
……
=[(b2)^2]^[2^(n-2)]=(b2)^[2^(n-1)]
=[(b1)^2]^[2^(n-1)]=(b1)^(2^n)
=(2√2-3)^(2^n)
bn=(2√2-3)^[2^(n-1)]
(an-√2)/(an+√2)=bn=(2√2-3)^[2^(n-1)]
an-√2={(2√2-3)^[2^(n-1)]}(an+√2)
={(2√2-3)^[2^(n-1)]}an+{(2√2-3)^[2^(n-1)]}√2
{1-(2√2-3)^[2^(n-1)]}an=√2{1+(2√2-3)^[2^(n-1)]}
an=√2{1+(2√2-3)^[2^(n-1)]}/{1-(2√2-3)^[2^(n-1)]}
什么意思 a(n+1)是an后一项吧 把an≤a 改为an≤t
我会回来的。。。。。。