4. Let $p(x)$ be a polynomial of degree $2n$, $p(0)=p(2)=\cdots=p(2n)=0, p(1)=p(3)=\cdots=$ $p(2n-1)=2, p(2n+1)=-30$. Find $n$.
4. From $\Delta^{2 n+1} p(0)=\sum_{b=0}^{2 n+1}(-1)^{k} \cdot C_{2 n+1}^{k} \cdot p(2 n+1-k)=0$, we have $(-30)+\sum_{j=1}^{n} C_{2 n+1}^{2 j} \cdot 2=0$, which simplifies to $(-30)+2 \cdot\left(\frac{1}{2} \cdot 2^{2 n+1}-1\right)=0$, solving for $n$ yields $n=2$.