# Degree of secant varieties of Veronese varieties

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 $$1 < k < n$$?

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 $$n+1-k$$.
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 $$2^n$$.