Example 5 Arrange all powers of 3 and the sums of distinct powers of 3 in an increasing sequence:
$$
1,3,4,9,10,12,13, \cdots \text {. }
$$

Find the 100th term of this sequence.
Analysis: Write the known sequence in the form of the sum of powers of 3:
$$
\begin{array}{l}
a_{1}=3^{0}, a_{2}=3^{1}, a_{3}=3^{1}+3^{0}, \\
a_{4}=3^{2}, a_{5}=3^{2}+3^{0}, a_{6}=3^{2}+3^{1}, \\
a_{7}=3^{2}+3^{1}+3^{0}, \cdots
\end{array}
$$

It is easy to find that the terms correspond exactly to the binary representation of the natural number sequence:
$$
\begin{array}{l}
1=2^{0}, 2=2^{1}, 3=2^{1}+2^{0}, \\
4=2^{2}, 5=2^{2}+2^{0}, 6=2^{2}+2^{1}, \\
7=2^{2}+2^{1}+2^{0}, \cdots
\end{array}
$$

Since $100=(1100100)_{2}=2^{6}+2^{5}+2^{2}$,
the 100th term of the original sequence is
$$
3^{6}+3^{5}+3^{2}=981 \text {. }
$$

Explanation: This example demonstrates the superiority of base $p$ $(p \neq 0)$. Generally, each base system has its strengths and weaknesses. Depending on the specific problem, flexibly choosing the base system often yields surprising results. Please see another example.