= Solution
The initial values are $\beta_0(t)=1$ and $\beta_1(t)=0$. Repeatedly applying the <Brownian moment recursion> to odd indices reduces them to $\beta_1$, so every odd moment vanishes. For even indices, if the formula holds at $2k-2$, then
$$
\begin{aligned}
\beta_{2k}(t)&=k(2k-1)\int_0^t\frac{(2k-2)!s^{k-1}}{2^{k-1}(k-1)!}\,ds\\
&=\frac{(2k)!}{2^kk!}t^k.
\end{aligned}
$$
The base case $k=0$ is $\beta_0=1$. Thus
$$
\boxed{\beta_{2k+1}(t)=0,\qquad\beta_{2k}(t)=\frac{(2k)!t^k}{2^kk!}\qquad(k\ge0).}
$$
Equivalently the even moments are $(2k-1)!!\,t^k$, the familiar Gaussian moment formula with <variance> $t$.
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