4. The increasing sequence $1,3,4,9,10,12,13, \cdots$, consists of some positive integers, which are either powers of 3, or the sum of several different powers of 3. The 100th term of this sequence is ().
A. 729
B. 972
C. 243
1). 981
4. D

Because the terms of this sequence are of the form: $k_{0} \cdot 3^{0}+k_{1} \cdot 3^{1}+k_{2} \cdot 3^{2}+\cdots+k_{m} \cdot 3^{m}$ (where each $k_{i} \in\{0,1\}$). The first 6 powers of 3, $3^{0}, 3^{1}, 3^{2}, \cdots, 3^{5}$, can form $2^{6}-1=63$ different sums of powers of 3, which are the first 63 terms of the sequence; the 64th term is $3^{6}=729$, starting from the 65th term, there are $C_{5}^{1}+C_{5}^{2}+\cdots+C_{5}^{5}=2^{5}-1=31$ terms that do not contain $3^{5}=243$, the 96th term should be $729+$ $243=972$, followed by $729+243+1,729+243+3,729+243+1+3$, thus the 100th term is $729+243+9=981$.