Prove that $\sum_{i=0}^{50} {50 \choose i}^2 = {100 \choose 50}$

Question

Prove that $$\sum_{i=0}^{50} {50 \choose i}^2 = {100 \choose 50}$$

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I tried writing the following but did not get anywhere.

$$ {50 \choose 0} {50 \choose50} + {50 \choose 1} {50 \choose49} + \dots + {50 \choose 50} {50 \choose 0} $$

I even thought about pairing the $n$-th term and the respective $51$st term and taking $2$ common (except for $50 \choose 25$, which is alone). I am too scared to take the LCM. Is there any special property that I am unaware of?

To be honest, I don't have any idea from where to even start.

Answer 1

A quick solution by double counting can be given as noted in the above comment. Another standard algebraic approach (which is the same in essence) but it is a cute idea to use polynomials for those kinds of things is the following.

Consider the polynomial:

$$(x+1)^{100}$$

In view of Newton the coefficient of $x^{50}$ when expanding the above polynomial is $\binom{100}{50}$ . On the other hand you can expand it in a different way, one may write:

$$ (x+1)^{100} = (x+1)^{50}(x+1)^{50} $$

And expand after.Then the coefficient of $x^{50}$ by polynomial multiplication and using the binomial must be:

$$ \sum_{i=0}^{50} \binom{50}{i} \binom{50}{50-i} = \sum_{i=0}^{50} \binom{50}{i}^2 $$

but these two coefficients must be the same and the claim follows.