$ \begin{aligned} (3)解:原式&=(2-1)×(1+2+2^{2}+2+···+2^{2023}) \\ &= 2^{2024}-1. \\ \end{aligned}$
$(4)解：设1-3+3^{2}-3^{3}+···-3^{49}+3^{50}=k，①$
$则①×3，得3-3^{2}+3^{3}-3^{4}+···-3^{50}+3^{51}=3k，②$
$①+②，得4k=3^{51}+1，所以k=\dfrac{3^{51}+1}{4}，$
$故原式的值为\dfrac{3^{51}+1}{4}.$
