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 $\qquad$
The terms in the sequence have the form $a_{0}+a_{1} \cdot 3+a_{2} \cdot 3^{2}+\cdots+a_{k} \cdot 3^{k}$, where $a_{k}=1, a_{i} \in\{0,1\}, i=0,1, \cdots$, $k-1$, so there are $2^{k}$ terms of this form. Since $1+2+\cdots+2^{5}=2^{6}-1=63$, the coefficients corresponding to the 64th term are $(0,0,0,0,0,0,1)$, the coefficients corresponding to the 96th term are $(0,0,0,0,0,1,1)$, and the coefficients corresponding to the 100th term are $(0,0,1,0,0,1,1)$.
Therefore, the 100th term of this sequence is $3^{6}+3^{5}+3^{2}=981$.