Example 5 (4th American Invitational Mathematics Examination) The increasing sequence $1,3,4,9,10,12,13, \cdots$ consists of some positive integers, which are either powers of 3 or sums of several different powers of 3. Find the 100th term of this sequence.
Solve: Write the known numbers in the form of powers of 3:
$$
\begin{array}{l}
\left.a_{1}=3^{0}, a_{2}=\right\lrcorner, a_{3}=3^{1}+3^{0}, a_{4}=3^{2}, a_{5}=3^{2}+3^{0}, a_{6}=3^{2}+3^{1}, \\
a_{7}=3^{2}+3^{1}+3^{0}, \cdots
\end{array}
$$

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 $3^{6}+3^{5}+3^{2}=981$.