$ x_n = 1 + 2a + 3a ^ 2 + \cdots + na ^ {n-1}$ has any three consecutive terms coprime Let $ a \in \mathbb N $. For $ n \in \mathbb N $, let $$ x_n = 1 + 2a + 3a ^ 2 + \cdots + na ^ {n-1}. $$
$ x_1 = 1, x_2 = 1 + 2a, \ldots $.
For $ n \in \mathbb N $, show that $ \gcd (x_n, x_ {n + 1}, x_ {n + 2}) = 1 $.
Any help, suggestion on how to start solving that exercise.
 A: Notice that $x_{n+2}-(a+1)x_{n+1}+ax_{n}=a^{n+1}$ so $gcd(x_{n},x_{n+1},x_{n+2})$ divides $a^{n+1}$, so if it is not 1 , say $m$ , $1<m<a^{n+1}+1, m \in \mathbb N,$ then since $x_{n+2}$ is equivalent to 1 modulo a , we have that $x_{n+2}$ is coprime to all prime divisors of $a$, so $m$ is coprime to $x_{n+2}$ since the prime divisors of $m$ are prime divisors of $a$ , so $m$ is not a divisor of $x_{n+2}$ so it is not the $gcd(x_{n},x_{n+1},x_{n+2})$, a contradiction.
A: Start by looking at  $x_{n+1} - x_n$ and $x_{n+2} - x_{n+1}$.
A: Hint: put $\,c_k = k\,$ below, which uses the Euclidean algorithm / Euclid's Lemma / gcd laws.
If $\ x_n = c_1 + c_2 a + \cdots + c_n a^{n-1}\ $ and $\ \color{#0a0}{(c_1,a)=1}\,$ and $\ \color{#c00}{(c_j,c_{j+1})=1\,\ {\rm for\ all}\,\ j}\:$ then
$$\begin{align} (x_n,x_{n+1},x_{n+2}) 
&= (x_n,\,x_{n+1}-x_n,\,x_{n+2}-x_{n+1})\\[.1em]
&= (x_n,\ \ \  c_{n+1} a^n,\ \ \ \  c_{n+2}a^{n+1})\\[.1em]
&= (x_n,\ \ \ c_{n+1},\ \ \ \ \ \ \ \ c_{n+2})\ \ {\rm by}\ \ (x_n,a) = \color{#0a0}{(c_1,a) = 1}\\[.1em]
&=\, 1,\ \  {\rm by}\ \ \color{#c00}{(c_{n+1}, c_{n+2}) = 1}
\end{align}\qquad$$
