3

Is there a closed form expression for $\displaystyle\sum_{k=0}^n \binom{2n}{2k}$?

A student I tutor was asking me about this and I didn't know. I know if this had a $k$ instead of $2k$, this sum would just be $(2n)^n$ but summing over evens complicates things. Is there even an asymptotic form for this?

Kashif
  • 1,497

4 Answers4

5

Hint: Use the binomial theorem to compute $$ 0^{2n}=(1-1)^{2n} $$ and $$ 2^{2n}=(1+1)^{2n} $$

robjohn
  • 345,667
  • Oh, I just did this using generating functions and got that as well. I just wonder if there's a good counting argument for why this identity holds. – Kashif May 08 '14 at 20:22
  • @ClarkKent: it's one of those things that once you see it, you'll never forget it :-) – robjohn May 08 '14 at 20:28
3

The set of evensized subsets of $\{1,\ldots,2n\}$ is in bijection with the power set of $\{1,\ldots,2n-1\}$: In one direction just drop $2n$ if necessary, in the other direction do or don't add $2n$ to ensure correct parity. This bijection shows that $$\sum_{k=0}^n{2n\choose 2k}=2^{2n-1}. $$

Hagen von Eitzen
  • 374,180
2

Since $\binom{2n}{k}$ is the number of subsets of $\{1,\ldots,2n\}$ with exactly $k$-elements, $$\sum_{k=0}^{n} \binom{2n}{2k} = \binom{2n}{0} + \binom{2n}{2} + \binom{2n}{4} + \ldots + \binom{2n}{2n}$$ is the number of subsets of $\{1,2,\ldots,2n\}$ which have an even number of elements.

For any nonempty, finite set $X$, exactly half the subsets of $X$ have even cardinality, and the other half have odd cardinality. To see this, fix some element $p \in X$. Define a function $f : \mathscr{P}(X) \to \mathscr{P}(X)$ by

$$ f(A) = \begin{cases} A \cup \{p\} & \text{ if } p \notin A \\ A \setminus \{p\} & \text{ if } p \in A \\ \end{cases} $$

The effect of $f$ is to toggle whether or not $p$ belongs to a subset. Note $f$ is a bijection. In fact, $f \circ f = \mathrm{id}$, so $f$ is its own inverse. Also, the cardinality of $f(A)$ is always $1$ away from the cardinality of $A$. So, $f$ maps the even-cardinality subsets to the odd-cardinality subsets, and vice versa.

In conclusion, one has $$ \sum_{k=0}^{n} \binom{2n}{2k} = \frac{\text{the number of subsets of $\{1,2,\ldots,2n\}$}}{2} = \frac{2^{2n}}{2} = 2^{2n-1}$$

Mike F
  • 22,196
  • High school me would have known this cold :|. – Kashif May 08 '14 at 22:42
  • @ClarkKent: Is your pupil a highschool student? – Mike F May 08 '14 at 22:45
  • He's a college student. I just used to be extremely good at combinatorics when I was in high school thanks to math olympiads. – Kashif May 08 '14 at 22:56
  • @ClarkKent: well, here's a (maybe?) fun follow-up problem for you: For what fraction of the $3^n$ maps $f : {1,\ldots,n} \to {1,2,3}$ is $\sum_{k=1}^n f(k)$ a multiple of $3$? – Mike F May 08 '14 at 23:35
1

Let me add another way of obtaining the same answer based on Pascal's Triangle. As you know, the sum of all elements of the row $a=2^a$. Take consecutive elements of row $2n-1$ and pair them off. The sum of each pair is equal to $\binom{2n}k$ for some odd $k$. The sum of $\binom{2n}k$ for all odd $k$ is equal to the sum of all these pairs, the sum of all elements of row $2n-1$, $2^{2n-1}$. The sum of $\binom{2n}k$ for all even $k$ then is simply the sum of all elements of row $2n$ minus the sum for all odd $k$.

$$2^{2n}-2^{2n-1}=2^{2n-1}$$

Mike
  • 13,318