Summation simplification $\sum_{k=0}^{n} \binom{2n}{k}^2$ $\sum_{k=0}^{n} \binom{2n}{k}^2$
So i'm trying to simplify this one and I'm stuck in nowhere. Some kind of tip would be appreciated.
Thanks! :)
 A: Try this:
$$\sum_{k = 0}^n \binom{2n}{k}^2 = \sum_{k = 0}^n \binom{2n}{k} \binom{2n}{2n - k} = \frac{1}{2} \sum_{k = 0}^{2n} \binom{2n}{k} \binom{2n}{2n - k} + \frac{1}{2}\binom{2n}{n}^2.$$
Then the sum turns into the coefficient described by anon in the answer below: the $x^{2n}$ term in the product
$$(1 + x)^{2n} (1 + x)^{2n} = (1 + x)^{4n}$$
which is $\binom{4n}{2n}$.  So the answer is
$$\frac{1}{2}\binom{4n}{2n} + \frac{1}{2} \binom{2n}{n}^2.$$
A: It suffices to find the summation indexed from $k=0$ to $k=2n$. This will be the constant coefficient of $(1+x)^{2n}(1+x^{-1})^{2n}=(1+x)^{4n}x^{-2n}$, or the $x^{2n}$ coefficient of $(1+x)^{4n}$.
A: Count the cardinality of $\mathcal A=\left\{(A,B):A\subseteq\left\{1,2,\ldots,2n\right\},B\subseteq\left\{2n+1,2n+2,\ldots,4n\right\},|A|=|B|\right\}$ in two ways.
Way 1: For each $k \in \left\{0,1,\ldots 2n\right\}$ choose $k$ elements for $A$ and $k$ elements for $B$. $$\displaystyle|\mathcal A|=\sum_{k=0}^{2n}{2n\choose k}{2n\choose k}=2\sum_{k=0}^{n}{2n\choose k}{2n\choose k}-{2n\choose n}^2$$
Way 2: For any $S\subseteq\left\{1,2,\ldots,4n\right\}$ with $|S|=2n$ take $A=S\cap\left\{1,2,\ldots,2n\right\}$ and $B=S^c\cap\left\{2n+1,2n+2,\ldots,4n\right\}$ where $S^c=\left\{1,2,\ldots,4n\right\}\setminus S$. Therefore $$\left|\mathcal A\right| = \left|\left\{S\subseteq\left\{1,2,\ldots,4n\right\}:|S|=2n\right\}\right|={4n\choose 2n}$$ 
A: Your sum $S=\sum_{i=0}^n\binom{2n}i^2$ would be easier to do if it summed all the way to $i=2n$. So write
$$
  S'=\sum_{i=0}^{2n}\binom{2n}i^2=\sum_{i=0}^n\binom{2n}i^2+\sum_{i=n+1}^{2n}\binom{2n}{2n-i}^2=2S-\binom{2n}n^2.
$$
Now
$$
  S'=\sum_{i=0}^{2n}\binom{2n}i\binom{2n}{2n-i}=\binom{4n}{2n}
$$
by the Vandermonde identity, and it follows that
$$
  S=\frac12\left(\binom{4n}{2n}+\binom{2n}n^2\right).
$$
A: Using the identities
$$
\sum_{k=0}^p\binom{n}{k}\binom{m}{p-k}=\binom{n+m}{p}
$$
and
$$
\binom{n}{n-k}=\binom{n}{k}
$$
and mapping $n,m,p\mapsto2n$, we get
$$
\begin{align}
\binom{4n}{2n}
&=\sum_{k=0}^{2n}\binom{2n}{k}^2\\
&=\sum_{k=0}^n\binom{2n}{k}^2+\sum_{k=n+1}^{2n}\binom{2n}{k}^2\\
&=\sum_{k=0}^n\binom{2n}{k}^2+\sum_{k=0}^{n-1}\binom{2n}{2n-k}^2\\
&=\sum_{k=0}^n\binom{2n}{k}^2+\sum_{k=0}^{n-1}\binom{2n}{k}^2\\
&=2\sum_{k=0}^n\binom{2n}{k}^2-\binom{2n}{n}^2\\
\sum_{k=0}^n\binom{2n}{k}^2&=\frac12\binom{4n}{2n}+\frac12\binom{2n}{n}^2
\end{align}
$$
A: Here is a closed from for the sum
$$\sum_{k=0}^{n} \binom{2n}{k}^2= \frac{1}{4}\,{\frac { \left( 2\,n+1 \right)  \left( {4}^{n}\pi \,\Gamma 
 \left( 2\,n+ \frac{1}{2} \right) \Gamma\left( n + 1\right)+{16}^{n} \left( 
\Gamma\left( n+\frac{1}{2} \right)  \right) ^{3} \right) }{\pi \,\Gamma 
 \left( n+\frac{3}{2} \right)  \left( \Gamma  \left( n+1 \right)\right)^{2}}}.$$
You can use Zeilberger's algorithm to find a closed form for the sum. 
Added: We can have the above formula in terms of binomials
$$\sum_{k=0}^{n} \binom{2n}{k}^2= \frac{{16}^{n}}{2} \left({2\,n-\frac{1}{2}\choose -\frac{1}{2}}+{n-\frac{1}{2}
\choose -\frac{1}{2} }^{2}\right)$$ 
A: There have been many good answers. Here is a "visual" result.
The sums of the squares of the binomial-coefficients occur in the diagonal, if you multiply the Pascal-matrix by its transpose. Let
$$ P = \begin{bmatrix} 
 1 & . & . & . & . & . \\
 1 & 1 & . & . & . & . \\
 1 & 2 & 1 & . & . & . \\
 1 & 3 & 3 & 1 & . & . \\
 1 & 4 & 6 & 4 & 1 & . \\
 1 & 5 & 10 & 10 & 5 & 1
 \end{bmatrix}$$
Then $P \cdot P^\tau $ is
$$ Q = P \cdot P^\tau = \begin{bmatrix} 
 1 & 1 & 1 & 1 & 1 & 1 \\
 1 & 2 & 3 & 4 & 5 & 6 \\
 1 & 3 & 6 & 10 & 15 & 21 \\
 1 & 4 & 10 & 20 & 35 & 56 \\
 1 & 5 & 15 & 35 & 70 & 126 \\
 1 & 6 & 21 & 56 & 126 & 252
 \end{bmatrix}$$ 
(The part of proving this and providing the general formula is not done here).    
Then your question relates to the dot products of the even indexed rows/columns 
at the same index (beginning at zero) which are visibly 
in the diagonal: $1,6,70,\ldots$. And since the sums go only to the half 
of the index and the rows in the pascal-matrix are symmetric we arrive at the 
half of that values which is, after the index 0 $3,35,462,\ldots$ or 
$$ S_1(n) = {\binom{2(2n)}{2n} \over 2} $$ 
But at the even-indexed rows the Pascalmatrix has an odd number of terms, the middle term has been counted only once, so its half must again be added, so we arrive at
$$ S(n) = {\binom{2(2n)}{2n} \over 2}  + {\binom{2n}{n}^2 \over 2} $$ 
