Let $C\subset\mathbb{P}^n$ be a rational normal curve of degree $n$, and let $Sec\_k(C)\subset\mathbb{P}^n$ be its $k$-th secant variety. By Theorem 1.1 in this paper:


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we have that $Sec\_k(C)$ is normal and $Sing(Sec\_k(C)) = Sec\_{k-1}(C)$. Does $Sec\_k(C)$ have ordinary singularities of multiplicity two along $Sec\_{k-1}(C)\setminus Sec\_{k-2}(C)$?


More precisely let $f:X\rightarrow\mathbb{P}^n$ be the blow-up of $\mathbb{P}^n$ along $Sec\_{k-1}(C)$ with exceptional divisor $E$, and let $Y$ be the strict transform of $Sec\_k(C)$. Is the following statement true? The strict transform $Y$ is smooth, it intersects $E$ transversally and we have
$$Y = f^{\*}Sec\_k(C)-2E.$$
I guess this should be true for instance when we consider a rational normal curve $C$ of degree four and $Sec\_2(C)$ which is a cubic hypersurface. 



Singularities of secant varieties of rational normal curves
The following is a consequence of Theorem 1 in "A. Bertram, *Moduli of Rank-$2$ Vector Bundles, Theta divisors, and the geometry of curves in projective space*, J. Differential Geom. 35, 1992, 429-469."


Let $C\subset\mathbb{P}^{2h}$ be a degree $2h$ rational normal curve. Consider the following sequence of blow-ups:


* $\pi\_1:X\_1\rightarrow\mathbb{P}^{2h}$ the blow-up of $C$,
* $\pi\_2:X\_2\rightarrow X\_{1}$ the blow-up of the strict transform of $Sec\_2(C)$,


$\vdots$
* $\pi\_{h-1}:X\_{h-1}\rightarrow X\_{h-2}$ the blow-up of the strict transform of $Sec\_{h-1}(C)$.


Let $\pi:X\rightarrow\mathbb{P}^{2h}$ be the composition of these blow-ups. Then, for any $k\leq h$ the strict transform of $Sec\_{k-1}(C)$ is smooth, irreducible and transverse to all exceptional divisors. In particular $Y$ is smooth and the divisor in $Y$ given by the union of the exceptional divisors and the strict transform of $Sec\_{h}(C)$ is simple normal crossing.


It is enough to apply Theorem 1 of the above cited paper and observe that The rational normal curve is given by the Veronese embedding induced by the line bundle $L = \mathcal{O}\_{\mathbb{P}^1}(2h)$ on $\mathbb{P}^1$. Now,
$$h^{0}(\mathbb{P}^1,L(-2h)) = 1 = 2h+1-2h= h^{0}(\mathbb{P}^1,L)-2h.$$
This means that $C\subset\mathbb{P}^{2h}$ is embedded by a $2h$-very ample line bundle.


