Consider the degree two Veronese embedding $\mathbb{P}^n\rightarrow\mathbb{P}^N$ and let $V^n\_{2}\subset\mathbb{P}^N$ be the corresponding Veronese variety.


Let $Sec\_k(V^n\_{2})\subseteq\mathbb{P}^N$ be the $k$-secant variety of $V\_{2}^{n}$. This is the closure of the union of all $(k-1)$-planes spanned by $k$ independent points on $V\_2^n$.


Is there a closed formula for the degree of $Sec\_k(V^n\_{2})$?


For instance if $k = 1$ we have that $Sec\_1(V\_2^n) = V\_2^n$ has degree $2^n$, while for $k = n$ we have that $Sec\_n(V\_2^n)\subset\mathbb{P}^N$ is a hypersurface of degree $n+1$. What about the degree of $Sec\_k(V^n\_{2})\subseteq\mathbb{P}^N$ for $1 < k < n$?



Degree of secant varieties of Veronese varieties
The secant variety $Sec\_k(V^n\_2)$ is the variety parametrizing $(n+1)\times (n+1)$ symmetric matrices modulo scalar of rank at most $k$ that is of corank at least $n+1-k$.


Then by Proposition 12(b) in


J. Harris; L. W. Tu, *On symmetric and skew-symmetric determinantal varieties*, Topology 23 (1984), no. 1, 71–84.


the degree of $Sec\_k(V^n\_2)$ is given by


$$\deg(Sec\_k(V^n\_2)) = \prod\_{i=0}^{n-k}\frac{\binom{n+1+i}{n+1-k-i}}{\binom{2i+1}{i}}$$


In particular, for $k = n$ you get $n+1$, and for $k = 1$ you get $2^n$.


