4. Let the sequence $\left\{u_{n}\right\}$ be defined by $u_{1}=5, u_{n+1}-u_{n}=3+4(n-1)(n=1,2,3, \cdots)$. If $u_{n}$ is a polynomial in $n$, then the sum of the coefficients of this polynomial is
A. 3
B. 4
C. 5
D. 6
4. C

Solution 1 Given $u_{k+1}-u_{k}=3+4(k-1)$, let $k=n-1, n-2, \cdots 2,1$, and we can sequentially obtain
$$
\begin{array}{l}
u_{n}-u_{n-1}=3+4(n-2), \\
u_{n-1}-u_{n-2}=3+4(n-3) . \\
\cdots \cdots \\
u_{3}-u_{2}=3+4 \cdot 1, \\
u_{2}-u_{1}=3+4 \cdot 0 .
\end{array}
$$

Adding both sides of the above equations: the left side $=u_{n}-u_{1}$
the right side $=3 \cdot(n-1)+4[(n-2)+(n-3)+\cdots+1]=2 n^{2}-3 n+1$.
Given $u_{1}=5$, thus $u_{n}=2 n^{2}-3 n+1+5=2 n^{2}-3 n+6$.
Therefore, the sum of the coefficients is $2-3+6=5$. Hence, the answer is $\mathrm{C}$.
Solution 2 Let $u_{n}=a_{0}+a_{1} n+a_{2} n^{2}+\cdots+a_{k} n^{k}$,
from $u_{1}=a_{0}+a_{1}+a_{2}+\cdots+a_{k}$, and $u_{1}=5$, we get $a_{0}+a_{1}+a_{2}+\cdots+a_{k}=5$. Hence, the answer is C.