How do I prove the following?

$$\sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix}^2 = \begin{pmatrix} 2n \\ n \end{pmatrix}$$

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$$\sum_{r=0}^{2n}\binom{2n}{r}x^r = (1+x)^{2n} = (1+x)^n(1+x)^n$$

$$= [\sum_{k=0}^n \binom{n}{k}x^k]^2 = \sum_{k=0}^n \sum_{l = 0}^{n}\binom{n}{k} \binom{n}{l} x^{k+l}$$

Now this all seems very unrelated to the issue. But by equating the coefficients of \(x^r\) you can notice that: $$\binom{2n}{r} = \sum_{0 \le k,l \le n \ : \ k+l = r} \binom{n}{k} \binom{n}{l} = \sum_{k=0}^n \binom{n}{k} \binom{n}{r-k}$$

In the case where \(r=n\),

$$\binom{2n}{n} \sum_{k=0}^n \binom{n}{k} \binom{n}{n-k}=\sum_{k=0}^n \binom{n}{k} \binom{n}{k}$$

There is also a 'wordier' proof: Suppose there are \(2n\) members of parliament and exactly half belong to party \(A\) and half belong to party \(B\). How many ways are there to select a cabinet of \(r\) ministers? The answer is \(\binom{2n}{r}\). We can also arrive at this answer by considering the number of possibilities with the added constraint that \(k\) (out of \(r\)) ministers come from party \(A\), and then summing over \(k\):

$$\sum_{k=0}^r \binom{n}{k} \binom{n}{r-k} = \binom{2n}{r}$$

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First, notice that \( \binom{n}{k} = \binom{n}{n-k} \).

Using this identity one can rewrite the LHS as $$\sum_{k=0}^n \binom{n}{k}^2 = \sum_{k=0}^n \binom{n}{k}\binom{n}{n-k}$$

Now \( \bf{Vandermonde's Theorem} \) states that $$ \sum_{k=0}^r \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r} $$ where \(m,n,r \in \mathbb{N_0} \)

Applying this to \(\sum_{k=0}^n \binom{n}{k}\binom{n}{n-k}\) results in

$$\sum_{k=0}^n \binom{n}{k}^2 = \sum_{k=0}^n \binom{n}{k}\binom{n}{n-k}=\binom{2n}{n}$$

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Not sure why the hyperlink isn't working properly, since it works in the 'Preview' of my answer, so here it is: https://en.wikipedia.org/wiki/Vandermonde%27s_identity