A convolution identity $\sum_{k=0}^n {2k\choose k}\cdot{2(n-k)\choose n-k}=4^n $ for central binomial coefficients

Question

I have to prove that

$$\sum_{k=0}^n {2k\choose k}\cdot{2(n-k)\choose n-k}=4^n $$ I tried to do this by induction but I totally stucked. Is there any way that I can solve this equation maybe another method . Thanks for any help.

Answer 1

That is a consequence of the extended binomial theorem in the form $$\forall x\in(-1,1),\qquad \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\binom{2n}{n}\frac{x^n}{4^n} $$ just square both sides, then extract the coefficient of $[x^n$].
Have a look at page 15 here.

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Combinatorial interpretation: $T_n=\sum_{k=0}^{n}\binom{2k}{k}\binom{2n-2k}{n-k}$ is the number of strings with length $2n$ over $\Sigma=\{A,B,C,D\}$ with the following structure: $$ ABBAB\ldots BBAAADCDCC\ldots DDCCD $$ i.e. a prefix made by $A$s and $B$s only, a postfix made by $C$s and $D$s only, the same number of $A$s and $B$s, the same number of $C$s and $D$s. What happens if we consider a string in $T_{n+1}$ and we remove its last two characters? If we remove $CD,DC,BA$ or $AB$ we get a string in $T_n$. Vice-versa, by appending $CD$ or $DC$, or by pre-pending $AB$ or $BA$, to a string in $T_n$ we get a string in $T_{n+1}$. With a bit of extra work$^{(*)}$ it is not difficult to prove that $T_{n+1}=4\cdot T_n$ and to reach the same conclusion as above.

$(*)$ Indeed an explicit bijection between $T_n\times\{1,2,3,4\}$ and $T_{n+1}$ can be built by considering the first instant ensuring an equality between the number of $A$s and $B$s in the prefix / $C$s and $D$s in the postfix and by inserting/removing some string from $\{AB,BA,CD,DC\}$ there.