Evaluating the sum of $\sum_{k=0}^n \binom{n}{k}^2$ [duplicate]

I don't know if this question is trivial but let me put it in the first place. I'm trying to find the sum of $$\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+\cdots+\binom{n}{n-1}^2+\binom{n}{n}^2$$ or equivalently $$\displaystyle \sum_{k=0}^n \binom{n}{k}^2$$.

How would you proceed?

marked as duplicate by Holo, Hans Lundmark, user90369, Cesareo, Carl SchildkrautOct 10 '18 at 23:50

• What have you tried? – Jaroslaw Matlak Oct 10 '18 at 14:57
• You should not say that without providing a link, because I am pretty involved in mathematics and I did not know that. – HackerBoss Oct 10 '18 at 15:00
• Agreed. It is equal to $\binom{2n}n$, which can be proved using a combinatorial argument. – астон вілла олоф мэллбэрг Oct 10 '18 at 15:01
• It is $$\frac{2^{2 n} \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi } \Gamma (n+1)}$$ – Dr. Sonnhard Graubner Oct 10 '18 at 15:02
• It is a special case of Vandermonde's identity – Sil Oct 10 '18 at 15:03

lets first prove a more general thing, i.e,

$$\sum_{i = 0}^{x} {n \choose i } {m \choose x - i } = {m + n \choose x}$$.

For that, let's suppose we have m apples and n oranges, and now we want to choose x fruits from them, we can do it in $${m + n \choose x}$$ ways, and that's RHS. Now we can also select x fruits, by selecting 0 apples and x oranges, 1 apple and x - 1 oranges and so on, and that's exactly LHS, and hence LHS = RHS.

Now just put m = n, and x = n, in the equation, so it becomes

$$\sum_{i = 0}^{n} {n \choose i}{n \choose n - i} = {2n \choose n}$$

or,

$$\sum_{i = 0}^{n} {n \choose i} {n \choose i} = {2n \choose n}$$

$$\sum_{i = 0}^{n} {n \choose i}^2 = {2n \choose n}$$