If this is homework, than I greatly regret posting this answer. I expect you to say so immediately, and I will consider the repercussions later.
What you should do is consider summing the standard combination forms - for example, on 3 objects, we are counting the sum of the situations when you take: 0 of them, 1 of them, 2 of them, and all of them. But if we take 2 of them, we know there are $3 \choose 2$ different ways to do this. In general, choosing k of n objects can be done in $n \choose k$ - hence the semantic phrasing "n choose k."
So a naive answer would be to simply sum these. But you'll notice a pattern if you do this. In fact, I won't explicitly answer what this is, but instead I'll encourage you to try it out. But I'll explain why it's true.
I will assume that you are familiar with the binomial theorem and its relationship to combinations; then you also know that we are really considering the coefficients of $(x+y)^n$. If we want to sum the coefficients, there is no reason not to just let $x = y = 1$, so that the coefficients are always multiplied by 1 according to the theorem...
And from a little experimentation and the above paragraph, I think you will find what you need.
EDIT: While I like this answer because it gives a good understanding behind the combinatorial aspect of the question, one could also simply consider whether each element is in the set. So there are 2 possibilities for each element...
Note: If this is homework, please add that tag. I am changing your tags a little bit on this question, as I don't think it's a linear algebra question.