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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 $1 < k < n$?

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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 $2^n$.

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