Prove that if $a \mid b_1, \; a \mid b_2, \; \ldots$, and $a \mid b_n$ then $a \mid (b_1x_1 + b_2 x_2 + \ldots + b_n x_n)$ for any integers $x_1, x_2, \ldots , x_n$.
We'll show this by induction on $n$.
Let's take the statement $P(n)$ to mean
"if $a \mid b_1, \; a \mid b_2, \; \ldots, \;a \mid b_n$ then $a \mid (b_1x_1 + b_2 x_2 + \ldots + b_n x_n)$ for any integers $x_1, x_2, \ldots , x_n$"
Base Case
The base case is $P(0)$ which is vacuously true. This is a little abstract so let's take $P(1)$ as the base case, which is
"if $a \mid b_1$ then $a \mid (b_1x_1)$ for any integer $x_1$"
We need to show this is true. The assumption $a \mid b_1$ means, for some integer $k$
$$b_1 = a k$$
Multiplying both sides by $x_1$ we have
$$ b_1 x_1 = a (k x_1) $$
Which means $a \mid b_1 x_1$. So we have shown the base case $P(1)$ to be true.
Inductive Step
We need to show the inductive step $P(n) \implies P(n+1)$ is true.
To do this we assume $P(n)$ and derive $P(n+1)$.
The induction hypothesis, $P(n)$ is
"if $a \mid b_1, \; a \mid b_2, \; \ldots, \;a \mid b_n$ then $a \mid (b_1x_1 + b_2 x_2 + \ldots + b_n x_n)$ for any integers $x_1, x_2, \ldots , x_n$"
We need to show $P(n+1)$ which is
"if $a \mid b_1, \; a \mid b_2, \; \ldots, \;a \mid b_n, \; a \mid b_{n+1}$ then $a \mid (b_1x_1 + b_2 x_2 + \ldots + b_n x_n + b_{n+1} x_{n+1})$ for any integers $x_1, x_2, \ldots , x_{n+1}$"
Let' start with the expression
$$ \underbrace{ b_1x_1 + b_2 x_2 + \ldots + b_n x_n }_\textrm{A} \; + \; \underbrace {b_{n+1} x_{n+1}}_\textrm{B}$$
The induction hypothesis tells us the first part $A$ is divisible by $a$. The additional assumption that $a \mid b_{n+1}$ tells us the second part $B$ is also divisible by $a$. So the full expression is divisible by $a$.
So we've shown $P(n+1)$ to be true.
By induction, $P(n)$ is true for all natural numbers $n$, and so the given statement is true.