Combinatorial interpretation of a sum

Question

I would like to know if there exists a way to interpret this sum by a combinatorial argument $$\sum _ { k = 0 } ^ { n } \frac { ( - 4 ) ^ { k } k ! } { ( 2 k + 1 ) ! ( n - k ) ! } = \frac { 1 } { ( 2 n + 1 ) \cdot n ! }$$ As of now I have tried coming up with some inclusion exclusion argument by multiplying the sum on the LHS by $(2n+1)n!$ but have failed in doing so.

Would appreciate any guidance or help towards a solution

Edit: I found this identity from this mathematical reflections paper I came across https://drive.google.com/file/d/1Q_TdS1btxtMd-wkanMPGSHeXg3Wq2y-P/view?usp=drivesdk

Answer 1

Inclusion-exclusion will need $(-1)^k\binom{n}{k}$, so also multiply the summand by $k!/k!$ to rewrite as $$\sum_{k=0}^n (-1)^k\binom{n}{k} \frac{(2n+1)4^k}{(2k+1)\binom{2k}{k}}=1.$$