1

For $n\ge 0$, show $\sum_{k=0}^{n} \binom{2k}{k} \binom{2(n-k)}{n-k} = 4^n$.

Want to first clarify the logic of the problem.

It is not only having variable $k$ (in number of items to be chosen in each category) as variable, but also having the quantity available in each category as variable.

I mean if instead of taking variable quantity : $2k, 2(n-k)$, have fixed qty. (say, $x,y$), then can define a problem as choosing a total of $k$ items from two categories having fixed quantities $x,y$.

But, now the two input quantities are variable and this makes it difficult to find a problem that can be represented by the equality.

Now, try to prove it using MI.

Over a well ordered set of non-negative integers, induction can be used.

Base case: $n=0$
$\binom{0}{0} \binom{0}{0} = 1$.

Taking the rhs, get : $4^0= 1$.

Induction hypothesis: it is true for natural number $n$.

So, $\sum_{k=0}^{n} \binom{2k}{k} \binom{2(n-k)}{n-k} = 4^n$

Inductive step: need show for next natural $n+1$

$\sum_{k=0}^{n+1} \binom{2k}{k} \binom{2(n+1-k)}{n+1-k} = 4^{n+1}$

This should lead to :

$4^n + \binom{2(n+1)}{(n+1)} \binom{2((n+1)-(n+1))}{((n+1)-(n+1))}= 4^{n+1}$

$\implies \binom{2(n+1)}{(n+1)} = 4^{n}.(3)$

Please help to proceed further, or suggest alternative approach.

jiten
  • 4,524
  • Related but more difficult https://math.stackexchange.com/questions/3229256/proving-that-sum-k-02n-2k-choose-k-2n-choose-k-left-frac-12 – Albus Dumbledore Feb 03 '21 at 06:36
  • Note that the inductive step would be $$\sum_{k=0}^{n+1} \binom{2k}{k} \binom{2(n+1-k)}{n+1-k} = 4^{n+1}$$ which is more than simply adding a term to the previous sum. – Brian Moehring Feb 03 '21 at 06:42
  • As a sidenote, the top two answers in AlbusDumbledore's link use the generating function of the central binomial coefficient. Your problem is almost trivial if you know this, so do you know this generating function? (I doubt it, as I think your identity is how I derived this generating function way back when...) – Brian Moehring Feb 03 '21 at 06:47
  • @BrianMoehring please elaborate how going to 'inductive step' is more than adding a single term to the 'Induction hypothesis'. – jiten Feb 03 '21 at 06:50
  • 1
    Just write out the sum for some small values of $n$. It should be apparent that more is going on than simply adding a single term. – Brian Moehring Feb 03 '21 at 06:54
  • @BrianMoehring sorry, tried with $n=3$, but got no help in understanding. Though. got for $$k=0, \binom{0}{0}\binom{2.(3-0)}{3}=20$$ for $$k=1, \binom{2}{1}\binom{2.(3-1)}{3-1}=12$$ for $$k=2, \binom{2.2}{2}\binom{2.(3-2)}{3-2}=12$$ for $$k=3, \binom{2.3}{3}\binom{2.(3-3)}{3-3}=20$$ leading to sum of $64= 4^3$. Even couldn't still get understanding of type of problem representated. – jiten Feb 03 '21 at 07:05
  • 1
    In case it helps:

    $$\sum_{k=0}^{n} \binom{2k}{k} \binom{2(n-k)}{n-k} = \sum_{k=0}^{n} \frac{2k!}{(k!)^{2}} \frac{(2(n-k))!}{((n-k)!)^{2}} = \sum_{k=0}^{n}\frac{\binom{n}{k} \binom{n}{n-k} \binom{2n}{n}}{\binom{2n}{2k}} = \binom{2n}{n} \sum_{k=0}^{n}\frac{\binom{n}{k} \binom{n}{n-k} }{\binom{2n}{2k}}$$

     – open problem Feb 03 '21 at 07:41
  • 2
    @jiten: There are quite a few algebraic proofs of various sorts on this site. There are also at least two combinatorial proofs, one by Marc van Leeuwen here, and one by me here. – Brian M. Scott Feb 03 '21 at 07:48
  • @openproblem seems have taken first row from the post; but the other is different. There seems no link between the two rows. The second one has different no. of terms, and different terms too in numerator and denominator. – jiten Feb 03 '21 at 14:02
  • Does this answer your question? Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$ – metamorphy Apr 05 '22 at 11:57

1 Answers1

1

Take squares on both sides of the expansion $$(1-4x)^{-1/2}=\sum_{n=0}^{\infty}{2n \choose n}x^n.$$

WimC
  • 32,192
  • 2
  • 48
  • 88