Exercise 1551Combinatorics › Binomial coefficientsL1Added 3 Nov 2026

Sum of squares of binomial coefficients

#square sum#Vandermonde#binomial

Statement

Show that $\sum_{k=0}^{n}\binom{n}{k}^2=\binom{2n}{n}$ .

Hint

Apply Vandermonde's identity with $m=n$ and $k=n$ .

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Vandermonde. $\binom{m+n}{n}=\sum_{j=0}^{n}\binom{m}{j}\binom{n}{n-j}=\sum_j\binom{m}{j}\binom{n}{j}$ (using symmetry).

With $m=n$ . $\binom{2n}{n}=\sum_j\binom{n}{j}^2$ .

Conclusion. The identity holds.

Combinatorial proof. Choose $n$ from $2n$ by taking $j$ from the first $n$ and $n-j$ from the second.