Proof of the identity $\sum_{k=0}^{n}\binom {2k}{k}\binom {2(n-k)}{n-k}=4^n$

Question

Prove the following identity : $$\sum_{k=0}^{n}\binom {2k}{k}\binom {2(n-k)}{n-k}=4^n$$

I am sorry, this topic is very new to me and so is this website. I have looked at all the "duplicates" of this question on this website, but none have had an answer that I can comprehend. I thought I was able to just plug each one into the binomial theorem and arrive at the answer, but my professor stated that I am using the theorem incorrectly. I do not have an intuitive understanding of proofs yet. Any direction will help! Thank you.

Answer 1

I am helping you on this one. I am trying to make reference to this equality: $$\sum_{k=0}^n {{n}\choose{k}} = \sum_{k=0}^n {{n}\choose{k}} 1^k \cdot1^{n-k} = (1+1)^{k+n-k}=2^n$$ So, by the similar fashion (I will explain why it is the multiplicative) $$ \sum_{k=0}^n {{2k}\choose{k}} \cdot{{2(n-k)}\choose{n-k}} = \sum_{k=0}^n {{2k}\choose{k}}1^k1^k \cdot {{2(n-k)}\choose{n-k}} 1^{n-k}1^{n-k} = (1+1)^{k+k} \cdot (1+1)^{(n-k)+(n-k)} = 2^{2k+2n-2k}=2^{2n}=4^n$$ This actually leads to the good end. But I will establish why or improve the solution.

It seems like the following is problematic..... $$\sum_{k=0}^n {{2k}\choose{k}} \cdot{{2(n-k)}\choose{n-k}} = \sum_{k=0}^n \frac{(2k)!}{k!k!} \cdot \frac{[2(n-k)]!}{(n-k)!(n-k)!} = \sum_{k=0}^n \frac{(2n)!}{{{2n}\choose{2k}}} \cdot \Big( \frac{{{n}\choose{k}}}{n!} \Big)^2$$