$10 \cdot 55$ The sequence $1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2, \cdots$ has 1s separated by 2s, with the $n$-th pair of 1s separated by $n$ 2s. The sum of the first 1234 terms of the sequence is 
(A) 1996.
(B) 2419.
(C) 2429.
(D) 2439.
(E) 2449.
(47th American High School Mathematics Examination, 1996)
［Solution 1］Let the number of 1s in the first 1234 terms of the sequence be $k$. Thus, we have
$$
1234 \leqslant k+(1+2+3+\cdots+k)=k+\frac{k(k+1)}{2} \text {, }
$$

which simplifies to $k^{2}+3 k-2468 \geqslant 0$, yielding $k=49$.
$$
\therefore \quad S_{1234}=49+2(1234-49)=2419 \text {. }
$$

Therefore, the answer is $(B)$.
[Solution 2] Construct a complementary sequence $1,0,1,0,0,1,0,0,0,1,0,0,0,0,1,0, \cdots$ and add it term by term to the given sequence, resulting in $2,2,2,2, \cdots$
The number of 1s in the first 1234 terms of the complementary sequence (calculated in the same way as in Solution 1) is $k=49$, thus
$$
S_{1234}=1234 \times 2-49=2419 .
$$

Therefore, the answer is $(B)$.