Prove $2 \cdot \sum\limits_{k = 0}^{n} \binom{2n}{2k} = 2^{2n}$ without induction

Question

I am aware of $$ 2^{2n} = (1 + 1)^{2n} = \sum_{k = 0}^{2n} \binom{2n}{k} $$ I then tried grouping the terms and using that $\binom{n}{k} = \binom{n}{n-k}$, to obtain $$ \begin{align} \sum_{k = 0}^{2n} \binom{2n}{k} &= \binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{n - 1} + \binom{2n}{n} + \binom{2n}{n + 1} + \ldots + \binom{2n}{2n} \\ &= 2 \left[ \binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{n - 1}\right] + \binom{2n}{n} = \left( 2 \sum_{k = 0}^{n-1} \binom{2n}{k} \right) + \binom{2n}{n} \end{align} $$ But I don't know if I'm on the right track or how to progress from here.

score 3 · Accepted Answer · answered Mar 29 '18 at 00:10

3

I would also use $$0 = (1 - 1)^{2n} = \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}.$$ Adding this to the previous sum, $$2^{2n} = \sum_{k=0}^{2n}\left[(-1)^k\binom{2n}{k} + \binom{2n}{k}\right].$$ When $k$ is odd, the term is $0$, leaving only even $k$ terms (but doubled!). Therefore, $$2 \sum_{k=0}^{n} \binom{2n}{2k} = 2^{2n},$$ as required.

answered Mar 29 '18 at 00:10

Theo Bendit

50,900

1

Just an additional note about this technique: it's called a root-of-unity filter. Essentially, using the fact that $\sum_{z^n=1} z^k$ is $0$ if $n\nmid k$ and $n$ otherwise, one can derive lots of summation identities in this form summed only over multiples of $n$ (in this case, $n=2$, and we use the roots $z=1$ and $z=-1$). – Carl Schildkraut Mar 29 '18 at 23:47

Sri-Amirthan Theivendran · Answer 2 · 2018-03-29T00:16:57.640

3

Note that $$ \begin{align} (1+1)^{2n}&=\sum_{k=0}^{2n}\binom{2n}{k}\tag{1}\\ (1-1)^{2n}&=\sum_{k=0}^{2n}\binom{2n}{k}(-1)^k\tag{2} \end{align} $$ by the binomial theorem. Subtracting (2) from (1) we get that $$ 2^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}(1-(-1)^k)=2\sum_{0\leq k\leq2n, \, \text{k even}}\binom{2n}{k}=2\sum_{k=0}^{n}\binom{2n}{2k}. $$

edited Mar 29 '18 at 00:16

answered Mar 29 '18 at 00:11

Sri-Amirthan Theivendran

31,116

score 2 · Answer 3 · edited Mar 29 '18 at 00:35

The combinatorial proof does not always look out of place. Let us try it here.

The goal : number of subsets of even cardinality , of a set of $2n$ elements. Denote $[m] = \{1,2,...,m\}$.

There is a bijection $\phi$ between even subsets of $[m]$ and any subset of $[m-1]$ : $\phi(A) = A$ if $A$ has an even number of elements, and $\phi(A) = A \cup \{m\}$ otherwise. You can check this is a bijection. Therefore, the number of even subsets of $[m]$ is the number of subsets of $[m-1]$, which is $2^{m-1}$.
Also, every subset of $[2n]$ has size $2k$ for some $k \leq n$, so the number of such subsets is just the number of possible subsets for some fixed $k$, summed over all possible $k$. This leads to $\sum_{k=0}^n \binom{2n}{2k}$.

Setting $m = 2n$ in the first bullet point and equating the two explanations gives a combinatorial proof of the given identity.

Prove $2 \cdot \sum\limits_{k = 0}^{n} \binom{2n}{2k} = 2^{2n}$ without induction

3 Answers3