(3) The sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, \frac{a_{2 k}}{a_{2 k-1}}=2, \frac{a_{2 k+1}}{a_{2 k}}=3, k \geqslant 1$; then the sum of its first 100 terms $S_{100}=$ $\qquad$ .
$\begin{array}{c}3 \frac{3}{5}\left(6^{50}-1\right) \quad \text { Hint: } \frac{a_{2 k+1}}{a_{2 k-1}}=\frac{a_{2 k+1}}{a_{2 k}} \cdot \frac{a_{2 k}}{a_{2 k-1}}=6, \frac{a_{2 k+2}}{a_{2 k}}=\frac{a_{2 k+2}}{a_{2 k+1}} \cdot \\ \frac{a_{2 k+1}}{a_{2 k}}=6, a_{1}=1, a_{2}=2 \text {, so, } a_{2 k-1}=6^{k-1}, a_{2 k}=2 \cdot 6^{k-1}, \\ S_{100}=\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}\right)+\cdots+\left(a_{99}+a_{100}\right) \\ =3 \sum_{k=1}^{50} 6^{k-1}=\frac{3}{5}\left(6^{50}-1\right) .\end{array}$