33.41 In a non-decreasing sequence of odd numbers
$$
\left\{a_{1}, a_{2}, a_{3}, \cdots\right\}=\{1,3,3,3,5,5,5,5,5, \cdots\}
$$

each positive odd number \( k \) appears \( k \) times. It is known that there exist integers \( b, c, \) and \( d \) such that for all positive integers \( n \), \( a_{n}=b[\sqrt{n+c}]+d \), where \( [x] \) denotes the greatest integer less than or equal to \( x \). Then \( b+c+d \) equals
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(31st American High School Mathematics Examination, 1980)
[Solution] In the known general term formula, set $n=1,2,3,4,5$ to get
$$
\begin{array}{l}
1=b[\sqrt{1+c}]+d \\
3=b[\sqrt{2+c}]+d \\
3=b[\sqrt{3+c}]+d \\
3=b[\sqrt{4+c}]+d \\
5=b[\sqrt{5+c}]+d
\end{array}
$$

From (2) - (1) we have $2=b([\sqrt{2+c}]-[\sqrt{1+c}])$.
Since $0<[\sqrt{2+c}]-[\sqrt{1+c}]<2$, then $[\sqrt{2+c}]-[\sqrt{1+c}]=1$, and $b=2$.
From (3) - (2), (4) - (3), (5) - (4) we have
$$
\begin{aligned}
1+[\sqrt{1+c}] & =[\sqrt{2+c}]=[\sqrt{3+c}]=[\sqrt{4+c}] \\
& =[\sqrt{5+c}]-1 .
\end{aligned}
$$

From this, we know that there exist some integers $m$ and $k$ such that $2+c=k^{2}, 5+c=m^{2}$. Since the only perfect squares that differ by 3 are 1 and 4, hence $k=1$, and $m=2$.
Therefore, $c=-1, d=1$.
Conversely, substituting $b=2, c=-1, d=1$ into the original general term formula also holds true. Hence, the answer is $(C)$.