I have $$\sum_{k=0}^n \binom{n}{k} = 2^n$$
I am proving this using a story, but I was wondering if my story is correct.
Our story is going to be we choose $k$ people out of $n$ to be in a club.
LHS: Considering all calses, we choose $k$ people to form to club from $n$ people.
RHS: Out of $n$ people, they choose either to be a part of the club or not.
By this we show that the LHS and RHS count the same thing.
Anything I should clarify, or anything I have wrong. I am trying to work out the kinks for these types of proofs.
I think you're nearly there, there are just a few things which could probably be just a touch clearer.
Mainly, if we're counting the number of ways to pick $k$ people to join the club, clearly for any case where people are chosen to join the club there will be some number $k$ which are chosen, and that number will be between $0$ and $n$, and any given case will only ever have one value of $k$.
These couple of distinctions, while important, are what makes the underlying concept of the proof work because otherwise you can't justify using the sum on the left-hand side to add up the counts for each case to get the total number.
If there were cases which had no corresponding value of $k$, or whose value of $k$ wasn't between $0$ and $n$, then the sum on the left could be undercounting. Otherwise, if a given case could have more than one value for $k$, then you could be overcounting. As it is the different values of $k$ set up disjoint spanning subsets of the set of ways to select people to join the club, so you can justify adding up the sizes of the subsets to get the size of the total set.
tl;dr: Yes this works, if you want to get into a bit more of the technical underpinnings then basically this is the rule for the size of the union of disjoint sets.