For a sequence $u_1,u_2\dots$, define $\Delta^1(u_n)=u_{n+1}-u_n$ and, for all integer 
$k>1, \Delta^k(u_n)=\Delta^1(\Delta^{k-1}(u_n))$. 
If $u_n=n^3+n$, then $\Delta^k(u_n)=0$ for all $n$
$\textbf{(A) }\text{if }k=1\qquad \\ \textbf{(B) }\text{if }k=2,\text{ but not if }k=1\qquad \\ \textbf{(C) }\text{if }k=3,\text{ but not if }k=2\qquad \\ \textbf{(D) }\text{if }k=4,\text{ but not if }k=3\qquad\\ \textbf{(E) }\text{for no value of }k$
Note that
\begin{align*} \Delta^1(u_n) &= \left[(n+1)^3+(n+1)\right]-\left[n^3+n\right] \\ &= \left[n^3+3n^2+4n+2\right]-\left[n^3+n\right] \\ &= 3n^2+3n+2, \\ \Delta^2(u_n) &= \left[3(n+1)^2+3(n+1)+2\right]-\left[3n^2+3n+2\right] \\ &= \left[3n^2+9n+8\right]-\left[3n^2+3n+2\right] \\ &= 6n+6, \\ \Delta^3(u_n) &= \left[6(n+1)+6\right]-\left[6n+6\right] \\ &= \left[6n+12\right]-\left[6n+6\right] \\ &= 6, \\ \Delta^4(u_n) &= 6-6 \\ &= 0. \end{align*}
Therefore, the answer is $\boxed{\textbf{(D)}}.$
More generally, we have $\Delta^k(u_n)=0$ for all $n$ if $k\geq4.$
~MRENTHUSIASM