Consider the following series function $$ f_n (R) = \sum_{n=1}^{\infty} \frac{3\left(1-\Lambda\right)\left(2n+1\right)\left(\Lambda-n-2\right)}{4 \left(n+2-3\Lambda \right)} \, R^{2n-1} \, , $$ where $\Lambda,R \in [0,1)$.
For $\Lambda = 0$, one gets $$ f_n(R) = \frac{3}{4} \frac{R \left( R^2-3 \right)}{\left( 1-R^2\right)^2} \, . $$
For $\Lambda = 1/2$, one gets $$ f_n(R) = \frac{3}{8} \frac{R \left( 3R^2-5 \right)}{\left( 1-R^2\right)^2} \, . $$
But, is this only possible for those 2 special values? Is there a way to prove that somehow?
Thank you and stay safe and sound in view of what is going around us.
The general expression of $f(\Lambda,R)$ (your notation $f_n(R)$ looks strange) is possible in particular in terms of incomplete beta functions and Hurwitz-Lerch transcendent function.
$$-\frac{3}{4} (1-\Lambda )f(\Lambda,R)=$$ $$(2-\Lambda ) R \Phi \left(R^2,1,3-3 \Lambda \right)+R^{6 \Lambda -5} \left((7-2 \Lambda ) B_{R^2}(3-3 \Lambda ,-1)+4 B_{R^2}(4-3 \Lambda ,-2)\right)$$
Exploring , I only found "simple" explicit expressions for $\Lambda=\frac k 6$ $(k=0,1,2,3,4,5)$
$$\Lambda=0 \implies\frac{3 R \left(R^2-3\right)}{4 \left(R^2-1\right)^2}$$ $$\Lambda=\frac 1 6\implies\frac{5 \left(11 R^6-37 R^4+20 R^2-12\right)}{72 R^3 \left(R^2-1\right)^2}+\frac{5 \tanh ^{-1}(R)}{6 R^4}$$ $$ \Lambda=\frac 13\implies \frac{\left(R^2-2\right) \left(5 R^2+1\right)}{6 R \left(R^2-1\right)^2}-\frac{\log \left(1-R^2\right)}{3 R^3}$$ $$\Lambda=\frac 12\implies\frac{3 R \left(3 R^2-5\right)}{8 \left(R^2-1\right)^2}$$ $$\Lambda=\frac 23\implies \frac{R \left(11 R^2-17\right)}{12 \left(R^2-1\right)^2}+\frac{\log \left(1-R^2\right)}{3 R}$$ $$\Lambda=\frac 56\implies\frac{R \left(13 R^2-19\right)}{24 \left(R^2-1\right)^2}-\frac{5}{6} \tanh ^{-1}(R)$$
Obviously, there are many other but the formulae start to be quite messy. They would require some extensive search using a CAS (just as I did for the above).
For example , $$\frac{160}{9} R^{7/2} \left(R^2-1\right)^2\,f\left(\frac 14,R\right)=$$ $$2 \sqrt{R} \left(7 R^6-29 R^4+27 R^2-15\right) +15 \left(R^2-1\right)^2 \left(\tan ^{-1}\left(\sqrt{R}\right)+\tanh ^{-1}\left(\sqrt{R}\right)\right)$$ $$\frac{32}{3} \sqrt{R} \left(R^2-1\right)^2 \,f\left(\frac 34,R\right)=$$ $$4 R^{3/2} \left(2 R^2-3\right)+9 \left(R^2-1\right)^2 \left(\tan ^{-1}\left(\sqrt{R}\right)-\tanh ^{-1}\left(\sqrt{R}\right)\right)$$