How to prove this combinatory equation?

Question

I don't know how to prove this equation: $\sum_\limits{k = 0}^n \binom{2k}{k} (^{2n-2k}_{\ \ n-k}) = 4 ^ n.$ I read here Combinatorial proof that $\sum \limits_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$ when $n$ is even about possible proofs, but my teacher said me that I've to use $\binom{-1/2}{k}$ in my solution. How can I use it if I mustn't use infinite series?

Answer 1

EDIT: A proof using $\binom{-1/2}{k}$.

Note that $\binom{2k}{k}=(-4)^k\binom{-1/2}{k}$. Then using Vandermonde's identity, \begin{align*}\sum_{k=0}^n\binom{2k}{k}\binom{2n-2k}{n-k} &= (-4)^n \sum_{k=0}^n\binom{-1/2}{k}\binom{-1/2}{n-k} \\&= (-4)^n\binom{-1}{n} \\ &= 4^n.\end{align*}

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Here's a nice combinatorial proof.

Consider in the alphabet $\{A,B\}$ for which in any initial string of letters (a prefix), there are at least as many $A$'s as $B$'s. Call this type of word majorized, and call a word with exactly as many $A$'s as $B$'s balanced.

Lemma. The number of majorized words of length $2n$ in the alphabet $\{A,B\}$ is $\tbinom{2n}{n}$.

Proof. It suffices to establish a bijection between the majorized words and the balanced words, which number $\tbinom{2n}{n}$.

• For any word which has some prefix with more $B$'s than $A$'s, we can: find the prefix of minimal length for which the difference in the number of $B$'s and the number of $A$'s is maximal, and then change the last letter (necessarily a $B$, by maximality) into an $A$.
• Conversely, for any word with more $A$'s than $B$'s, or more generally for a word for which some right-hand string of letters (a suffix) has this property, we can: find the suffix of minimal length for which the difference in the number of $A$'s and the number of $B$'s is maximal, and then change the first letter (necessarily an $A$, by maximality) into a $B$.

Note that these steps are inverses, so given a balanced word, we can transform it using the first step into a majorized word, and conversely, given a majorized word of length $2n$ with $\delta \geq 0$ more letters $A$ than $B$, we can use the second operation $\tfrac 12 \delta$ times to obtain a balanced word, thus establishing the bijection. $\square$

We now count the number of words of length $2n+1$ with more letters $A$ than $B$. Clearly by symmetry, there are $\tfrac{1}{2} \times 2^{2n+1}=4^n$ such words.

Counting in a different way, consider the longest prefix (possibly null) which is balanced: if it has length $2k$, then there are $\tbinom{2k}{k}$ possibilities for this prefix: the next letter is then necessarily $A$ (from the maximality of the prefix), and the suffix is of length $2n-2k$. Treating this suffix as a separate subword, each prefix of this suffix is majorized, otherwise we would contradict maximality: by the lemma, there are $\tbinom{2n-2k}{n-k}$ such suffixes. Hence summing from $0 \leq k \leq n$, $$\sum_{k=0}^n\binom{2k} {k}\binom{2n-2k}{n-k}=4^n. $$

Answer 2

If you are aware that $$ \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\,x^n = \frac{1}{\sqrt{1-x}} \tag{A}$$ the proof is straightforward. By squaring both sides of $(A)$ we get $$ \sum_{n\geq 0}\frac{x^n}{4^n}\sum_{a+b=n}\binom{2a}{a}\binom{2b}{b} = \frac{1}{1-x} = \sum_{n\geq 0} x^n \tag{B} $$ and by comparison of coefficients: $$ \sum_{a+b=n}\binom{2a}{a}\binom{2b}{b}=4^n.\tag{C}$$