# Summation of even combinations

In the post Sum of combinations of n taken k where k is from n to (n/2)+1, it has been explained clearly how to calculate the summation of combinations from n/2 to n. I was wondering if there is any such formula for calculating the summation of odd/even combinations if n is even/odd.

For example,
Formula for summation of all possible combinations of n = 2^n

Suppose, if n is odd,
Summation of odd combinations = summation of even combinations = 2^(n-1).

How to calculate the summation of even or odd combinations if n is even? I am trying to come up with some kind of formula but couldn't. Any help would be appreciated.

$$2^n = (1+1)^n = \displaystyle\sum_{k = 0}^{n}\dbinom{n}{k} = \sum_{\substack{k = 0 \\ k \ \text{is even}}}^{n}\dbinom{n}{k} + \sum_{\substack{k = 0 \\ k \ \text{is odd}}}^{n}\dbinom{n}{k}$$
$$0 = (1-1)^n = \displaystyle\sum_{k = 0}^{n}\dbinom{n}{k}(-1)^k = \sum_{\substack{k = 0 \\ k \ \text{is even}}}^{n}\dbinom{n}{k} - \sum_{\substack{k = 0 \\ k \ \text{is odd}}}^{n}\dbinom{n}{k}$$
Now, you just have to solve for $\displaystyle\sum_{\substack{k = 0 \\ k \ \text{is even}}}^{n}\dbinom{n}{k}$ and $\displaystyle\sum_{\substack{k = 0 \\ k \ \text{is odd}}}^{n}\dbinom{n}{k}$
Hint: You can split into two cases according to whether a fixed element is in the combination or not, and thus reduce even/odd counting formulas for even $n$ to formulas for the case of odd $n-1$.