Is there a closed form formula for this recurrence relationship? I wondered whether $A_{n+1}= 100A_n^2 + 200A_n$ can be expressed in closed form? Ie when the next term in the recurrent relationship is composed of some constant time the square of the previous term PLUS some other constant times the previous term. 
Thanks in advance. 
 A: This is a tricky problem in general, and I don’t know of any nice closed form representation for the recurrence you’ve brought up:
$$A_{n+1}=100A_n^2+200A_n$$
However, regarding your more general question, there are some special cases in which the recurrence
$$a_{n+1}=\alpha a_n^2+\beta a_n$$
can be solved in closed form. For instance, if $\beta=0$, the solution to the recurrence is
$$a_n=\alpha^{2^n-1}\cdot a_0^{2^n}$$
in terms of $a_0$, the initial term. Another special case occurs when $\beta=2$. In this case, the general solution to the recurrence is given by
$$a_n=\frac{(\alpha a_0+1)^{2^n}-1}{\alpha}$$
Another more obscure case can be solved using trigonometric functions. If $\beta^2/4\alpha-\beta/2\alpha=1$, then we have that
$$a_n=\frac{2}{\alpha}\cos(2^n\arccos(\alpha a_0/2+\beta/4))-\frac{\beta}{2\alpha}$$
...but this only applies when the domain for $\arccos$ contains $\alpha a_0/2+\beta/4$. However, we can use inverse hyperbolic cosine instead of inverse cosine if the argument goes outside of the domain of inverse cosine.
