let $\{a_{n}\}$ such $a_{1}=0,a_{n}<0,\forall n\ge 2$,such $$(a_{n+1}+1)^2(2a_{n+1}+1)=2a^4_{n+1}(2a_{n}+1)(a_{n}+1)$$
find the closed form $a_{n}$
it easy to get $$a_{2}=1-\sqrt{2},~~~~a_{3}=3+2\sqrt{2}-\sqrt{2(10+7\sqrt{2})}$$
also by condtion we get $$\dfrac{a_{n+1}+1}{a_{n}+1}\cdot\dfrac{2a_{n+1}+1}{2a_{n}+1}=\dfrac{2a^4_{n+1}}{a_{n+1}+1}$$
A closed form for $a_n$ is given by $$\tag1 a_n=-\frac{\cos\left(\frac\pi{2^n}\right)}{\cos\left(\frac\pi{2^n}\right)+1}\ \ .$$ The formula will be derived in three steps.
Step 1. $a_{n+1}=:x$ is defined as the unique negative solution of $$2(c+1)(2c+1)x^4-2x^3-5x^2-4x-1=0,\mbox{ where }c=a_n.$$ Now there is a factorisation $$\tag2 2(c+1)(2c+1)x^4-2x^3-5x^2-4x-1=\\\hspace{10em} [(2c+1)x^2-2x-1][(2c+2)x^2+2x+1],$$ as can be easily verified. This will be used to show inductively
Lemma 1 The sequence $a_n$ is well defined and we have $-\frac12<a_n\leq0$ for all $n$.
For $n=0$, this is true. If it is true for some $n$ then we have to show that $2(c+1)(2c+1)x^4-2x^3-5x^2-4x-1=0,c=a_n$, has a unique negative solution $x$ and that $-\frac12<x\leq0$.
Using (2), any solution of $2(c+1)(2c+1)x^4-2x^3-5x^2-4x-1=0$ satisfies $(2c+1)x^2-2x-1=0$ or $(2c+2)x^2+2x+1=0.$ Now the discriminant of the latter equation is $-4-8c<0$ because $c>-\frac12$. Hence the latter equation has no real solution. Clearly the former equation has a unique negative solution. Now the value of $(2c+1)x^2-2x-1$ at $x=0$ is $-1$ whereas at $x=-\frac12$ it is $\frac14(2c+1)>0$. Hence the zero is strictly between $-\frac12$ and 0. From $(2c+1)x^2-2x-1=0$ we find that $$\tag3 (2a_n+1)a_{n+1}^2-2a_{n+1}-1=0\mbox{ or } a_{n+1}=\frac{1-\sqrt{2a_n+2}}{2a_n+1}.$$
Step 2. The formulas (3) remind the double- and half-angle formulas for the trigonometric and hyperbolic functions. We would like to reduce (3) to a pair of those by a change of variables. More precisely, we want to reduce $$\tag4 a_n=\frac1{a_{n+1}}+\frac1{2a_{n+1}^2}-\frac12$$ to some double angle formula by putting $a_n=M(A_n), a_{n+1}=M(A_{n+1})$ with some Möbius transformation $$M(z)=\frac{rz+s}{tz+u}\mbox{ with certain constants }r,s,t,u.$$ Calling the right hand side of (4) $f(a_{n+1})$, observe that the fixed points of $f$ in the extended complex plane are the solutions of $(2w+1)w^2=2w+1$ and hence $w=-1,-\frac12$ and $w=1$. Furthermore the derivatives $f'(w)$ are 0,4 and $-2$ at $w=-1,-\frac12$ and $w=1$, respectively. These values do not change by a conjugation $f\mapsto M^{-1}fM$. Comparing with the properties of the correponding maps in the double angle formulas, we find a match in $\cos(2\phi)=2\cos^2(\phi)-1=:g(\cos(\phi))$: The fixed points of $g$ are $v=\infty,-\frac12,1$ and the derivatives are $0,-2,4$, respectively. This suggests that we use $M$ exchanging $-\frac12$ with 1 and $-1$ with $\infty$, respectively. A small calculation yields $$M(z)=M^{-1}(z)=-\frac z{z+1}.$$ Another small calculation yields $g=MfM$ or more explicitely $$A_{n}=2A_{n+1}^2-1\mbox{ if }A_n=M(a_n), A_{n+1}=M(a_{n+1})$$ as desired. Observe that we only have to consider nonnegative $A_n$, because $M$ maps $]-\frac12,0]$ to $[0,1[$. Hence $$\tag5 A_{n+1}=\sqrt{\tfrac12(A_n+1)}.$$
Step 3. Given $a_1=0$, we have $A_1=0$ and we can write $A_1=\cos(\frac\pi2)$. Since (5) is the half angle formula for $\cos$, we find using mathematical induction that $$A_n=\cos\left(\frac\pi{2^n}\right)\mbox{ for all }n.$$ As $a_n=M(A_n)$ for all $n$, we finally obtain that $$a_n=-\frac{\cos\left(\frac\pi{2^n}\right)} {\cos\left(\frac\pi{2^n}\right)+1}\mbox{ for all }n,$$ that is formula (1). This completes the proof.
