$12 \cdot 11$ If $n$ is a multiple of 4, and $i$ is the imaginary unit, then the sum $S=1+2 i+3 i^{2}$ $+\cdots+(n+1) i^{n}$ equals
(A) $1+i$.
(B) $\frac{1}{2}(n+2)$.
(C) $\frac{1}{2}(n+2-n i)$.
(D) $\frac{1}{2}[(n+1)(1-i)+2]$.
(E) $\frac{1}{8}\left(n^{2}+8-4 n i\right)$.
(15th American High School Mathematics Examination, 1964)
[Solution] Consider $S-i S=\sum_{k=0}^{n} i^{k}-(n+1) i^{n+1}$, that is,
$$
(1-i) S=\frac{1-i^{n+1}}{1-i}-(n+1) i^{n+1}=1-(n+1) i \text {, }
$$

Note that $n$ is a multiple of 4, so $i^{n}=1$. Then
$$
S=\frac{1-(n+1) i}{1-i}=\frac{1}{2}(n+2-n i) .
$$

Therefore, the answer is $(C)$.