Prove that$$\overbrace{\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2}} + \overbrace{\binom n {\frac{n+1} 2} + \cdots + \binom n {n-1} + \binom n n} = 2^n. $$
Background:
This problem came from the following link:
https://math.stackexchange.com/posts/1639757/edit
and I didn't understand why that was true in which no one on the page has proved it to be so. and i was wondering what was the proof on that
People have already mentioned the algebraic proof in the comments. Here is a simple combinatorial proof:
Note that $\binom{n}{k}$ is the number of subsets of size $k$ of a set of size $n$. Thus, $\sum_{k=0}^n\binom{n}{k}$ is the total number of subsets of a set of size $n$. But a set of size $n$ has $2^n$ substs, so $\sum_{k=0}^n\binom{n}{k}=2^n$.
Using proof by induction and some thinking using elementary set theory:
Theorem: $\sum_{k=0}^{n}{\binom{n}{k}} = 2^n$
Proof (Roughly): Our base case is simple. $\binom{0}{0} = 1 = 2^0 $
However we want to think about this along the lines of sets and their power sets. If we have a set $A$ with $n$ elements, then by definition the power set $P$ of $A$ is the set of all subsets of $A$.
Example: $A = \{1,2\}$ with $n=2$ elements. Then $P(A) = \{\emptyset,\{1\},\{2\},\{1,2\}\}$ with $2^n = 2^2 = 4$ elements or in context of the proof $\binom{2}{0}+\binom{2}{1}+\binom{2}{2} = 2^2$.
Now, in the spirit of induction I want to consider the set $A'$ that has one more element added to it. What happens to its power set? Well, using $A$ from before and adding our new element to be $3$ we get $A' = \{1,2,3\}$ with $n=3$ elements. Then $P(A') = \{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$ with $2^n = 2^3 = 8$ elements or $\binom{3}{0}+\binom{3}{1}+\binom{3}{2}+\binom{3}{3} = 2^3$.
In a more general sense, all we did was double the number of subsets of $A$ by adding our new element to all $4$ previous subsets, which gains us $4$ new subsets $+$ the $4$ old subsets. This is completely logical since the new set $A'$ is made only by adding 1 element to it, and so it should still retain the same subsets as $A$, with some more.
This is a concrete example using specific sets, however the logic of the proof is the same.
