$11 \cdot 21$ Arrange all powers of 3 and sums of distinct powers of 3 in an increasing sequence: $1,3,4,9,10,12,13, \cdots$ Find the 100th term of this sequence.
(4th American Mathematical Invitational, 1986)
[Solution] Each term of this sequence is a positive number, and its ternary representation consists only of 0s and 1s. Therefore, a one-to-one correspondence can be established between all positive integers and the terms of the sequence, such that the binary representation of number $n$ matches the ternary representation of the corresponding term in the sequence:
$$
\begin{array}{l}
1=.1_{(2)} \longleftrightarrow 11_{(3)}=1, \\
2=10_{(2)} \longleftrightarrow \quad 10_{(3)}=3, \\
3=11_{(2)} \longleftrightarrow 11_{(3)}=4, \\
4=100_{(2)} \longleftrightarrow 100_{(3)}=9, \\
5=101_{(2)} \longleftrightarrow 101_{(3)}=10, \\
6=110_{(2)} \longleftrightarrow 110_{(3)}=12, \\
7=111_{(2)} \longleftrightarrow \quad 111_{(3)}=13 .
\end{array}
$$

In this correspondence, the $k$-th positive integer corresponds to the $k$-th term of the sequence.
Therefore, to find the 100th term of the given sequence, we need to find the binary representation of the decimal number 100, and then find the ternary representation of this binary number.
$$
100=1100100_{(2)} \longleftrightarrow 1100100_{(3)}=981 .
$$

Thus, the 100th term of the sequence is 981 (in decimal).