Prove combinatorially $\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}=2^{n-1}$

Question

$$\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}=2^{n-1}$$

I have no idea how to start, can i get hints?

Answer 1

$$\binom{n}{0}+\binom{n}{2}+\binom{n}{4} \ldots$$

counts the number of even-sized subsets of $\{1 \ldots n\}$. I assume that you know the total number of subsets of $\{1\ldots n\}$ is $2^n$. To show the even-sized subsets number $2^{n-1}$, a direct way would be to show that it's the same as the number of subsets of $\{1 \ldots n-1\}$.

So: lay out all of your subsets of even size. For each set $S$, consider whether it contains $n$ or not. If it does not contain $n$, then it is an even sized subset of $\{1 \ldots n-1\}$. If it does contain $n$, then $S -\{n\}$ is an odd-sized subset of $\{1 \ldots n-1\}$. Clearly this is a reversible function - given any $T \subset \{1 \ldots n-1\}$, I can tell you the unique even subset of $\{1 \ldots n\}$ that maps to it (depending on whether it has even or odd size itself). It follows that the two sets have the same size.

Answer 2

Hint: Consider the following bijection $$\varphi:\mathcal{P}([n])\longrightarrow \mathcal{P}([n]),$$ given by $$\varphi (X)=\begin{cases}X\cup \{n\}&\text{if }n\not \in X\\ X\setminus \{n\}& \text{Otherwise}\end{cases}.$$ Show that this is a bijection that maps even sized to odd sized sets and viceversa. Use binomial theorem to conclude(or involution principle if you know it).

Answer 3

Maybe you can try by induction?

As a starting point, compute it for $n=0, 1, 2$, just to verify than the relation is true for some base cases.

Than, you can assume the relation to be true for $n$ and show that, with this assumption, then, the relation holds also for $n+1$.

And you're done!

I would try this way! Let me know ;)

Answer 4

Hint: Using the binomial theorem, what is $ ( 1 + 1) ^n$?
What is $ ( 1 - 1) ^n$?

Answer 5

The LHS gives the number of subsets of even size of a set with cardinality $n$.

Now, there are a total of $2^n$ subsets, and clearly exactly half of them will have even cardinality, i.e. $2^n/2 = 2^{n-1}$

Since both sides count the same thing, we are done.

Answer 6

\begin{align} \sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k} &=\sum_{k=0, k\text{ even}}^{n}\binom{n}{k} =\sum_{k=0}^{n}\binom{n}{k}-\sum_{k=0, k\text{ odd}}^{n}\binom{n}{k}\\ &=2^n+\sum_{k=0, k\text{ odd}}^{n}(-1)^k\binom{n}{k}\\ &=2^n+\sum_{k=0}^{n}(-1)^k\binom{n}{k}-\sum_{k=0, k\text{ even}}^{n}(-1)^k\binom{n}{k}\\ &=2^n+(1-1)^n-\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}\\ \implies 2\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k} &= 2^n \implies \sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k} = 2^{n-1} \end{align}