Help on proof of $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$

Question

The proof is required to be made through the binomial theorem. I will expose the demonstration I was tought, and forward my questions after exposing it. You'll see question marks like this one (?-n) on points I don't quite understand, where $n$ is the numeration of the mark. This are the doubts I have about the demonstration, the which I hope someone can clarify.

Prove that $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$.

We will use the following equality, and call it $P$:

$(1+x)^n(1+x)^n=(1+x)^{2n}$ (?-1)

The result will be proved finding the $x^n$ coefficient of both terms of this equality (?-2).

According to the binomial theorem, the left-hand side of this equation is the product of two factors, both equal to

$\binom{n}{0}1+\binom{n}{1}x+...+\binom{n}{r}x^r+...+\binom{n}{n}x^n$

When both factors multiply, a term on $x^n$ is obtained when a term of the first factor has some $x^i$ and the term of the second factor has some $x^{n-i}$. Therefor the coefficients of $x^n$ are

$\binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\binom{n}{2}\binom{n}{n-2}+...\binom{n}{n}\binom{n}{0}$ .

Since $\binom{n}{n-r}=\binom{n}{r}$, the previous summation is equal to $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2$. So the left hand side of the equation we are asked to proove is a coefficient of $x^n$. When we expand the right-hand side of the equation $P$, we find that $\binom{2n}{n}$ is a coefficient of $x^n$. Therefore (?-3) the left-hand side of the equation we were asked to prove is in deed equal to $\binom{2n}{n}$. In conclussion,

$\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$.

This was all the demonstration. My doubt one (?-1) goes about where the heck does this equation come from? How would I know what equation to come up with if requested to prove a different equality?

Doubt two (?-2) goes about why would the solution of the first equation would have anything to do with finding the $x^n$ coefficients of the one I just made up (see doubt one).

Doubt three (?-3) goes about why demonstrating that $a$ is a coefficient of $x^n$ on the left hand side of the equation I made up, and that $b$ is a coefficient of $x^n$ on the right-hand side of this equation as well, would prove my original equation, the one I was supposed to prove on the first place?

I know there are many doubts here, I hope you guys can help me. Sorry for the long post, it's a long demonstration.

Answer 1

Doubt 1: it isn´t an equation, but an identity (!!!): $${{a}^{n}}\cdot {{a}^{n}}\equiv {{\left( {{a}^{n}} \right)}^{2}}\equiv {{a}^{2n}},$$ or $${{a}^{n}}\cdot {{a}^{n}}\equiv {{a}^{n+n}}\equiv {{a}^{2n}}.$$Yes, Mathematics are indisputably important, and for many, also beautiful. But much more important than Mathematics and humility and gratitude.

In the brief indication that I have given you about your doubt number 1, there is the answer to all your doubts. However, it seems that my correction has bothered you.

Your algebraic development is correct and the only thing you need is to understand the validity of the arguments used. And all this is based on distinguishing between equation and identity: an equation is an equality that is verified only for certain values ​​of the variables, or for none, while an identity is an equality that is fulfilled for any value of the variables in its definition domain.

Since the equality $(1 + x) ^ n (1 + x) ^n = (1 + x) ^ {2n}$ is an identity, and two polynomials are identical (they take the same value for any value that is assigned to their variable) if and only if the coefficients of the terms of equal degree are identical, the coefficient of $x ^ n$ is the same in the development of $(1 + x) ^ n (1 + x) ^n$ than in that of $(1 + x) ^ {2n}$.

This simple reasoning is the idea underlying the validity of the argument used in the demonstration, that is, proving that the coefficient of $x ^ n$ is the same in the development of $(1 + x) ^ n (1 + x) ^n$ than in that of $(1 + x) ^ {2n}$.

Therefore, the answer given in my previous post was not trivial; It contained the key to solving all your doubts about the validity of your demonstration: identity is unconditional equality and in the case of the identity of polynomials, this unconditionality implies the equality of its coefficients in terms of the same degree. Keep in mind that a polynomial can be defined by omitting its variable, simply by the succession of its coefficients, so that two polynomials are identical if the sequences of their coefficients are equal.

Robert Shore (https://math.stackexchange.com/users/640080/robert-shore), Help on proof of $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$, URL (version: 2019-04-17): https://math.stackexchange.com/q/3191796 gives you the same clarifications.

Answer 2

You can think of Questions $1$ and $2$ together. You know that $\binom{2n}{n}$ is the coefficient of $x^n$ in $(1+x)^{2n}$, so with some experience it's obvious to guess that what you want to find is another useful expression that allows you to calculate that coefficient in a different way.

Question $3$ seems to result from a misunderstanding on your part. We're not talking about a coefficient of $x^n$. The left hand side is a polynomial in $x$, so it has an $x^n$ term which has a coefficient. The right hand side is also a polynomial in $x$, and the two polynomials are equal, so the coefficients of each and every power of $x$ (including $x^n$) have to be equal.

Calculating the same quantity (in this case, the coefficient of $x^n$) in two different ways ($(1+x)^n(1+x)^n$ and $(1+x)^{2n}$) is a common technique in mathematical proofs (particularly combinatorial proofs) for showing that two apparently different expressions are in fact equal.

Answer 3

To address (?-1):

This is closely related to what we call the generating function method, where a sequence $\{a_n\}$ is related with a (formal) polynomial $$A(x)=a_0x^0+a_1x^1+a_2x^2\dots.$$ We can see that multiplying two of these functions add as $A(x)+B(x)\sim \{a_n+b_n\}$, and they multiply as $A(x)B(x)\sim\{\sum_{i=0}^n a_ib_{n-i}\}.$ In this case, after your transformation $\binom{k}{n}\mapsto \binom{n-k}{n}$, there is an expression of the form $\sum_{i=0}^n a_ib_{n-i}$. So we try to use the multiplication of generating functions.