= Solution
Differentiating the Rodrigues identity gives $H_0=1$ and $H_{n+1}=2tH_n-H_n'$. Induct with the proposed finite sum. The coefficient at $t^{n+1-2j}$ in this recurrence is the contribution from $2tH_n$ minus the derivative contribution with previous index $j-1$:
$$
\frac{n!(-1)^j2^{n+1-2j}}{j!(n-2j)!}-\frac{n!(-1)^{j-1}2^{n-2j+2}}{(j-1)!(n-2j+1)!}.
$$
Using a common denominator, this equals $(n+1)!(-1)^j2^{n+1-2j}/[j!(n+1-2j)!]$. At an endpoint omit terms whose factorial indices would be negative; the same formula then covers the sole remaining term. Thus
$$
\boxed{H_n(t)=n!\sum_{j=0}^{\lfloor n/2\rfloor}\frac{(-1)^j2^{n-2j}}{j!(n-2j)!}t^{n-2j}.}
$$
The initial value and recurrence identify these polynomials with the given Rodrigues definition.
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