6. Given an increasing sequence composed of powers of 3 or the sum of several different powers of 3: $1,3,4,9,10,12,13, \cdots$. Then the 100th term of this sequence is $\qquad$
6. 981.

The terms of the sequence are given by $\sum_{i=0}^{n} 3^{i} a_{i}$, where,
$$
a_{i} \in\{0,1\}(i=1,2, \cdots, n) \text {. }
$$

When $n=5$, there are $2^{6}-1=63$ numbers that can be formed, and 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 $729+243=972$, and the 100th term is $972+9=981$.