Infinitely many primes of the $\sum_{i=0}^{a} n^{i}+m\cdot\sum_{i=0}^{b} n^{i}$ form? How to show that there is infinitely many prime numbers of the form:
$p=\sum_{i=0}^{a} n^{i}+m\cdot\sum_{i=0}^{b} n^{i}$
where: $m\in \mathbb{Z}^{*}$ , $a,b,n\in \mathbb{N}$ , $\gcd(a+1,b+1)=1$
For example: 
$\sum_{i=0}^{6} 10^{335\cdot i}+8648\cdot \sum_{i=0}^{4} 10^{335\cdot i}$ is prime number.
 A: I think that I have figured out answer on this question so here it is :
If we apply formula for the sum of the numbers in a geometric progression we may write :
$\sum_{i=0}^{a} n^{i}=\frac{n^{a+1}-1}{n-1}$ and $\sum_{i=0}^{b} n^{i}=\frac{n^{b+1}-1}{n-1}$ . Therefore we have to show that sequence :
$p_m=\frac{n^{a+1}-1}{n-1}+m\cdot \frac{n^{b+1}-1}{n-1}$ , contains infinitely many primes for any $a$ and $b$ .
According to Dirichlet's theorem there are infinitely many prime numbers in sequence :
 $a+n\cdot d$ , for any $(a,d)$ pair if $\gcd(a,d)=1$ , so in order to prove that sequence $p_m$ contains infinitely many primes we have to show that :
$$\gcd(\frac{n^{a+1}-1}{n-1},\frac{n^{b+1}-1}{n-1})=1$$
There is a known theorem that states :
$$\gcd(a^n-1,a^m-1)=a^{\gcd(n,m)}-1$$
so we may write following :
$$\gcd(n^{a+1}-1,n^{b+1}-1)=n^{\gcd(a+1,b+1)}-1=n-1 \Rightarrow$$
$$\Rightarrow \gcd(\frac{n^{a+1}-1}{n-1},\frac{n^{b+1}-1}{n-1})=1$$
The last equality implies that sequence :
$p_m=\sum_{i=0}^{a} n^{i}+m\cdot\sum_{i=0}^{b} n^{i}$ ,contains infinitely many prime numbers for any fixed $(a,b,n)$ triple,so there are infinitely many primes of the form : $\sum_{i=0}^{a} n^{i}+m\cdot\sum_{i=0}^{b} n^{i}$
