In a recent failed-post about some partial sums with respect to the Central Binomial and Catalan number the formulas $$\sum_{k=0}^n\frac{4^k}{B_k}=\frac{4^{n+1}(2n+1)}{B_{n+1}}+\frac{1}{3}$$ $\sum_{k=0}^n\frac{4^k}{B_k}=\frac{4^{n+1}(2n+1)}{3 B_{n+1}}+\frac{1}{3}$$ $$\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{4^{n+1}(2n^2+5n+2)}{B_{n+1}}+\frac{1}{5}$$ $\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{4^{n+1}(2n^2+5n+2)}{5 B_{n+1}}+\frac{1}{5}$$ were mention, here in MO, and I forgot to ask, so let me do it now:
Is This just one instance of some broader well known pattern?
Here, consider $B_m={2m \choose m}$ and $C_m=\frac{B_m}{m+1}$ for those set of numbers.
