10.68 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 .
(E) 11 .
(20th American High School Mathematics Examination, 1969)
[Solution 1] From the given condition $u_{k+1}-u_{k}=3+4(k-1)$, let $k=n-1, \dot{n}-2$, $\cdots 2,1$, and we can sequentially obtain
$$
\begin{array}{l}
u_{n}-u_{n-1}=3+4(n-2), \\
u_{n-1}-u_{n-2}=3+4(n-3), \\
\cdots \cdots \cdots \cdots \\
u_{3}-u_{2}=3+4 \cdot 1, \\
u_{2}-u_{1}=3+4 \cdot 0 .
\end{array}
$$

Adding both sides of the above equations:
$$
\begin{aligned}
\text { Left side } & =u_{n}-u_{1} \\
\text { Right side } & =3 \cdot(n-1)+4[(n-2)+(n-3)+\cdots+1] \\
& =2 n^{2}-3 n+1 .
\end{aligned}
$$

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 (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 $u_{1}=5$.
Thus, $a_{0}+a_{1}+a_{2}+\cdots+a_{k}=5$.
Hence, the answer is (C).