Credo sia la seconda: per verificarlo ti basta utilizzare la seguente espressione integrale dei polinomi di Hermite:
\begin{equation}
H_n(y) = \frac{e^{y^2}}{\sqrt{\pi}} \int_{-\infty}^{+\infty} (-2iu)^n e^{-u^2+2iu y} du
\end{equation}
Qua sotto ti posto la soluzione, in caso ti servisse per controllare:
Testo nascosto, fai click qui per vederlo
\begin{equation*}
\begin{aligned}
\sum_{k=0}^n \binom{n}{k} H_{n-k}(y)(2x)^k &= \sum_{k=0}^n \binom{n}{k} (2x)^k \frac{e^{y^2}}{\sqrt{\pi}} \int_{-\infty}^{+\infty} (-2iu)^{n-k} e^{-u^2+2iu y} du\\
&= \frac{e^{y^2}}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \sum_{k=0}^n \binom{n}{k} (2x)^k (-2iu)^{n-k} e^{-u^2+2iu y} du\\
&= \frac{e^{y^2}}{\sqrt{\pi}} \int_{-\infty}^{+\infty} (2x -2iu)^n e^{-u^2+2iu y} du\\
&= \frac{e^{y^2}}{\sqrt{\pi}} \int_{-\infty}^{+\infty} (-2iu)^n \exp\bigg[-\bigg(u + \frac{2x}{2i}\bigg)^2+2i\bigg(u + \frac{2x}{2i}\bigg) y\bigg] du\\
&= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{+\infty} (-2iu)^n \exp\bigg[-\bigg(u -ix-iy\bigg)^2\bigg] du\\
&= \frac{e^{(x+y)^2}}{\sqrt{\pi}} \int_{-\infty}^{+\infty} (-2iu)^n e^{-u^2 -2i(x+y)} du\\
&= H_n(x+y)
\end{aligned}
\end{equation*}