$17 \cdot 129$ An equilateral triangle has a side length of $a$. A new equilateral triangle is formed by connecting the midpoints of the sides of the original triangle, and a third equilateral triangle is formed by connecting the midpoints of the sides of the second triangle, and so on. The limit of the sum of the perimeters of these triangles is
(A) Indeterminate.
(B) $5 \frac{1}{4} a$.
(C) $2 a$.
(D) $6 a$.
(E) $4 \frac{1}{2} a$.
(2nd American High School Mathematics Examination, 1951)
[Solution] According to the problem, the sum of the perimeters of these triangles is
$$
\begin{aligned}
& 3 a+\frac{1}{2} \cdot(3 a)+\frac{1}{2^{2}} \cdot(3 a)+\cdots \\
= & 3 a\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots\right)=3 a \cdot \frac{1}{1-\frac{1}{2}}=6 a .
\end{aligned}
$$

Therefore, the answer is $(D)$.