5. Given the increasing sequence $1,3,4,9,10,12,13, \cdots$ where each term is either a power of 3 or the sum of several different powers of 3. Then the 100th term of this sequence is $\qquad$ .
5.981.

The terms of this sequence are
$$
a_{0}+a_{1} 3+a_{2} 3^{2}+\cdots+a_{n} 3^{n},
$$

where $a_{i} \in\{0,1\}, i=1,2, \cdots, n$.
When $n=5$, there can be $2^{6}-1=63$ numbers, the 64th term is $3^{6}=729$. Starting from the 65th term, there are $2^{5}-1=31$ terms that do not contain $3^{5}=$ 243, the 96th term is $729+$ $243=972$, followed by $729+243+1,729+243+3$, $729+243+1+3$, the 100th term is $729+243+9=$ 981.