1. (14 points) Let $n$ be a positive integer greater than 16, and let $A(n, k)$ and $B(n, k) (k \geqslant 3)$ be the number of $k$-term monotonic arithmetic sequences and geometric sequences with integer common ratios, respectively, in the set $\{1,2, \cdots, n\}$.
(1) Find the value of $A(2009,50)$;
(2) Prove: $n-4 \sqrt{n}+2<B(n, 3)<2 n-2 \sqrt{n}$.
1. (1) First, find the general $A(n, \dot{k})$.
Let the set $\{1,2, \cdots, n\}$ have a monotonically increasing arithmetic sequence $\left\{a_{i}\right\}$, with common difference $d$, and $a_{k}=a_{1}+(k-1) d$. Then
$$
\begin{array}{l}
1+(k-1) d \leqslant a_{1}+(k-1) d \leqslant n \\
\Rightarrow d \leqslant \frac{n-1}{k-1} .
\end{array}
$$

Let $p=\left[\frac{n-1}{k-1}\right]$.
Now, let $d \in\{1,2, \cdots, p\}$, and $a_{d} \in\{1,2, \cdots, n\}$ be the largest first term with common difference $d$. Thus,
$$
a_{d}+(k-1) d \leqslant n \text {. }
$$

Therefore, $a_{d}$ is the number of monotonically increasing arithmetic sequences with common difference $d$ that satisfy the given conditions. Then
$$
\begin{array}{l}
A(n, k)=2 \sum_{d=1}^{p} a_{d}=2 \sum_{d=1}^{p}[n-(k-1) d] \\
=2 n p-p(p+1)(k-1) .
\end{array}
$$

For $n=2009, k=50$, we have
$$
p=40, A(2009,50)=80360 \text {. }
$$
(2) Similarly, let the set $\{1,2, \cdots, n\}$ have a monotonically increasing geometric sequence $\left\{b_{i}\right\}$, with common ratio $r(r \geqslant 2, r \in \mathbf{N})$, and $b_{k}=b_{1} r^{k-1}$. Then
$$
r^{k-1} \leqslant b_{1} r^{k-1} \leqslant n \Rightarrow r \leqslant \sqrt[k-1]{n} .
$$

Let $q=[\sqrt[k-1]{n}]$, and let $r \in\{2,3, \cdots, q\}$, and $b_{r} \in\{1,2, \cdots, n\}$ be the largest first term with common ratio $r$. Thus, $b_{r} r^{k-1} \leqslant n$. Therefore, $b_{r}$ is the number of monotonically increasing geometric sequences with common ratio $r$ that satisfy the given conditions. Then
$$
B(n, k)=2 \sum_{r=2}^{q} b_{r}=2 \sum_{r=2}^{q}\left[\frac{n}{r^{k-1}}\right] \text {. }
$$

When $k=3$, we have
$$
\sqrt{n}-12 \sum_{r=2}^{q}\left(\frac{n}{r^{2}}-1\right) \\
> 2 \sum_{r=2}^{q} \frac{n}{r(r+1)}-2(q-1) \\
= 2 n\left(\frac{1}{2}-\frac{1}{q+1}\right)-2 q+2 \\
> n-4 \sqrt{n}+2 .
\end{array}
$$

Therefore, the conclusion holds.