1.an+1＝2[a(n-1)＋1]
(an+1)/[a(n-1)＋1]=2
{an+1}等比
an+1=(1+1)2^(n-1)=2^n
an=2^n-1
2.a(n+1)+3 =2(an＋3)
[a(n+1)+3]/(an＋3)=2
an＋3=(2+3)2^(n-1)
=5*2^(n-1)
3.f(a1)=2a1/(1+a1)=a2
a2=2a/(a+1)
2a2/(1+a2)=a3
a3=[4a/(a+1)]/[(2a+a+1)/(a+1)]
=4a/(3a+1)
a4=8a/(7a+1)
猜测:an=2^(n-1)a/{[2^(n-1)-1]a+1}
我验证过了,应该对的