Example 2 Let $\left\{a_{n}\right\}$ be the sequence of all numbers in the set $\left\{2^{s}+2^{t} \mid 10 \leqslant s<t\right.$ and $s, t \in \mathbf{Z}\}$ arranged in ascending order, i.e., $\square$
$$
\begin{aligned}
& a_{1}=3, a_{2}=5, a_{3}=6, a_{4}=9, a_{5}=10, \\
a_{6}= & 12, \cdots .
\end{aligned}
$$

Arrange the terms of the sequence $\left\{a_{n}\right\}$ in a triangular number table according to the principle of smaller on top, larger on the bottom, and smaller on the left, larger on the right:
$$
\begin{array}{l}
3 \\
5 \quad 6 \\
\begin{array}{lll}
9 & 10 \quad 12
\end{array} \\
\text {... ... ... ... } \\
\end{array}
$$
(1) Write out the numbers in the fourth and fifth rows of this triangular number table;
(2) $a_{100}$.
(1) Arrange the above triangular number table into a right-angled triangular number table:
$$
\begin{array}{l}
a_{1}=2^{0}+2^{1} \\
a_{2}=2^{0}+2^{2} \quad a_{3}=2^{1}+2^{2} \\
a_{4}=2^{0}+2^{3} \quad a_{5}=2^{1}+2^{3} \quad a_{6}=2^{2}+2^{3} \\
a_{7}=2^{0}+2^{4} \quad a_{8}=2^{1}+2^{4} \quad a_{9}=2^{2}+2^{4} \quad a_{10}=2^{3}+2^{4} \\
\cdots \ldots .
\end{array}
$$

It is easy to see that the numbers in the fourth row are $17, 18, 20, 24$, and the numbers in the fifth row are $33, 34, 36, 40, 48$.
(2) Note that the exponential function $f(x)=2^{x}$ is an increasing function, and
$$
2^{0}+2^{k+1}>2^{k-1}+2^{k}\left(k \in \mathbf{N}_{+}\right),
$$

Thus, the $k$-th row is
$$
2^{0}+2^{k}, 2^{1}+2^{k}, 2^{2}+2^{k}, \cdots, 2^{k-1}+2^{k} \text {. }
$$

Assume $a_{100}$ is in the $k$-th row. Since $a_{\frac{k(k-1)}{2}+i}=2^{i-1}$
$+2^{k}$, we have, when $\frac{k(k-1)}{2}+1 \leqslant 1000
\end{array}\right. \\
\Leftrightarrow \frac{-1+\sqrt{793}}{2}<k \leqslant \frac{1+\sqrt{793}}{2} \text {. } \\
\end{array}
$$

Since $k \in \mathbf{N}_{+}$, we have,
$$
k=14, a_{100}=2^{14}+2^{8}=16640 \text {. }
$$