The second exterior power of a half-spin representation of Spin(2n+4) is isomorphic to a direct sum of a bunch of exterior powers of the vector representation. Thus: \[\sum_{k=0}^{\lfloor n/4\rfloor}{2n+4 \choose n-4k} = {2^{n+1} \choose 2}.\] Might be a special case of the series multisection identities for binomial coefficients: https://en.wikipedia.org/wiki/Binomial_coefficient#Multisections_of_sums For n=3 we obtain the celebrated identity (16 choose 2) = (10 choose 3).