3. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, \frac{a_{2 k}}{a_{2 k-1}}=2, \frac{a_{2 k+1}}{a_{2 k}}=3(k \geqslant 1) \text {. }
$$

Then the sum of its first 100 terms $S_{100}=$ $\qquad$
3. $\frac{3}{5}\left(6^{50}-1\right)$.

Notice, $\frac{a_{2 k+1}}{a_{2 k-1}}=\frac{a_{2 k+1}}{a_{2 k}} \cdot \frac{a_{2 k}}{a_{2 k-1}}=6$,
$$
\begin{array}{l}
\frac{a_{2 k+2}}{a_{2 k}}=\frac{a_{2 k+2}}{a_{2 k+1}} \cdot \frac{a_{2 k+1}}{a_{2 k}}=6, \\
a_{1}=1, a_{2}=2 .
\end{array}
$$

Therefore, $a_{2 k-1}=6^{k-1}, a_{2 k}=2 \times 6^{k-1}$.
$$
\begin{array}{l}
\text { Then } S_{100}=\sum_{k=1}^{50}\left(a_{2 k-1}+a_{2 k}\right) \\
=3 \sum_{k=1}^{50} 6^{k-1}=\frac{3}{5}\left(6^{50}-1\right) .
\end{array}
$$