Let $\{a_n\}$ be a geometric sequence whose initial term is $a_1$ and common ratio is $q$. Show that $$a_1\binom{n}{0}-a_2\binom{n}{1}+a_3\binom{n}{2}-a_4\binom{n}{3}+\cdots+(-1)^na_{n+1}\binom{n}{n}=a_1(1-q)^n$$
where $n$ is a positive integer.
Because $\{a_n\}$ is a geometric sequence, therefore $a_k = a_1q^{k-1}$, $k=2$, $3$, $\cdots$. It follows that $$\begin{align*} & a_1\binom{n}{0}-a_2\binom{n}{1}+a_3\binom{n}{2}-a_4\binom{n}{3}+\cdots+(-1)^na_{n+1}\binom{n}{n} \\ =\ & a_1\displaystyle\binom{n}{0}-a_1q\binom{n}{1}+a_1q^2\binom{n}{2}-a_1q^3\binom{n}{3}+\cdots+(-1)^na_{1}q^n\binom{n}{n} \\ =\ &a_1\displaystyle\left(\binom{n}{0}-q\binom{n}{1}+q^2\binom{n}{2}-q^3\binom{n}{3}+\cdots+(-1)^nq^n\binom{n}{n}\right) \\ =\ &a_1(1-q)^n\end{align*}$$