Difficult Recurrence

As stated in the comments, due to the Descartes circle theorem four mutually tangent circles satisfy the identity: $$ 2(\kappa_1^2+\kappa_2^2+\kappa_3^2+\kappa_4^2)=(\kappa_1+\kappa_2+\kappa_3+\kappa_4)^2$$ where $\kappa_i$ is the curvature of the circle $\Gamma_i$, i.e. the reciprocal of the radius.

By setting $\kappa_n=\frac{1}{R(n)}$, Vieta jumping gives: $$ \kappa_0=0,\kappa_1=4,\quad\kappa_{n+1} = 2\kappa_n -\kappa_{n-1}+4,$$ hence: $$R(n)=\frac{1}{2n(n+1)}$$ and: $$\sum_{n=1}^{+\infty}R(n)^2 = \frac{\pi^2-9}{12},$$ since: $$\frac{1}{n^2(n+1)^2}=\left(\frac{1}{n}-\frac{1}{n+1}\right)^2=\frac{1}{n^2}+\frac{1}{(n+1)^2}-\left(\frac{2}{n}-\frac{2}{n+1}\right),$$ so $\zeta(2)=\frac{\pi^2}{6}$ and a new rational number make their appearance.

I believe your recurrence should have the form $$R(n) = \frac{\left(1 - 2 \sum_{k=1}^{n-1} R(k)\right)^2}{4\left(1 - \sum_{k=1}^{\color{red}{n-1}} R(k)\right)}.$$ To solve this, let $S(n) = \sum_{k=1}^n R(k)$, and solve the above expression for $S(n-1)$: $$S(n-1) = \frac{1}{2}\left(1 - R(n) - \sqrt{R(n)(2+R(n))}\right),$$ where we take the smaller root since we require $S(0) = 0$ for $R(1) = \frac{1}{4}$. Then we observe that $S(n) - S(n-1) = R(n)$, or $$2R(n) = R(n) - R(n+1) + \sqrt{R(n)(2+R(n))} - \sqrt{R(n+1)(2+R(n+1))}.$$ Now solving this recurrence for $R(n+1)$ yields $$R(n+1) = \frac{R(n)}{1 + 2R(n) + 2 \sqrt{R(n)(2+R(n))}}.$$ It is not difficult to show that if $R(1) = \frac{1}{4}$, then we can solve this recurrence explicitly to get $$R(n) = \frac{1}{2n(n+1)}.$$