Question

2a(n+1)=an^2+2an,a1=1/4,求an的通项

Answer 1

两边同时+1,则

2a（n+1）+1=(an+1)^2

由此式递推,

2a(n+1)+1=[a(n-1)+1]^4=[a(n-2)+1]^8=……=[a1+1]^(2^n)

代入a1=1/4

则a(n+1)=[(5/4)^(2^n)-1]/2

再把n+1换成n n换成n-1就OK了