设数列{an}满足下列关系式a1=2a (a是不为0的常数)an =2a - a^2 / an-1 数列bn=1/(an

a1=2a,
a2=2a-a^2/a1=2a-a^2/(2a)=3a/2
a3=2a-a^2/(a2)=2a-a^2/(3a/2)=2a-2a/3=4a/3
a4=2a-a^2/a3=2a-a^2/(4a/3)=2a-3a/4=5a/4
可以猜测：
an=(n+1)a/n
n=1,2,3,4时均成立,设n=k s时成立,即
ak=(k+1)a/k
则当n=k+1时,
a(k+1)=2a-a^2/ak
=2a-a^2/[(k+1)a/k]
=2a-ka/(k+1)
=[2-k/(k+1)]a
=[(2k+2-k)/(k+1)]a
=(k+2)a/(k+1)
即n=k+1时成立.故
an=(n+1)a/n.;
an-a=(n+1)a/n-a=[(n+1)/n-1]a=a/n
bn=1/(an-a)=n/a
bn=n/a