I'm trying to find a combinatorial proof (and algebraic too but that's optional) of the following result:
Let $S = \{2k+1 | 0 < 2k+1 ≤ n+1\}$

$$2^n = \sum\_{k\in S} \binom{n+1}{k}$$

I was thinking for the algebraic proof of using the identity:

$$2^{n+1} = \sum\_{k=0}^n \binom{n+1}{k}$$
In the following way:

$$2^{n+1}+(1-1)^{n+1} = \sum\_{k=0}^n \binom{n+1}{k} + \sum\_{k=0}^n (-1)^{k}\binom{n+1}{k} = 2\sum\_{k\in S}^n \binom{n+1}{k} $$

And dividing both sides by 2 gives us:
$$2^n =\sum\_{k\in S}^n \binom{n+1}{k} $$

Is my algebraic proof correct? Is there any other one you can provide? Also I've made no progress towards a combinatorial argument pls help :)