$10 \cdot 18$ For the sequence $u_{1}, u_{2}, \cdots$, define $\Delta^{1}\left(u_{n}\right)=u_{n+1}-u_{n}$, and for all integers $k>1$, define $\Delta^{k}\left(u_{n}\right)=\Delta^{1}\left[\Delta^{k-1}\left(u_{n}\right)\right]$. If $u_{n}=n^{3}+n$, then the condition for $\Delta^{k}\left(u_{n}\right)=0$ for all $n$ is
(A) when $k=1$.
(B) when $k=2$, but not when $k=1$.
(C) when $k=3$, but not when $k=2$.
(D) when $k=4$, but not when $k=3$.
(E) for no value of $k$.
(27th American High School Mathematics Examination, 1976)
[Solution] For all positive integers $n$, we have
$$
\begin{aligned}
\Delta^{1}\left(u_{n}\right) & =\left[(n+1)^{3}+(n+1)\right]-\left(n^{3}+n\right)=3 n^{2}+3 n+2, \\
\Delta^{2}\left(u_{n}\right) & =\left[3(n+1)^{2}+3(n+1)+2\right]-\left(3 n^{2}+3 n+2\right) \\
& =6 n+6, \\
\Delta^{3}\left(u_{n}\right) & =[6(n+1)+6]-(6 n+6)=6, \\
\Delta^{4}\left(u_{n}\right) & =0 .
\end{aligned}
$$

Therefore, the answer is $(D)$.