6. Let $k$ be a positive integer. We set $a_{1}=1$ and, for every $n \geq 1, a_{n+1}=k\left(a_{1}+\cdots+a_{n}\right)+1$ (so for example $a_{2}=k a_{1}+1$ and $a_{3}=k\left(a_{1}+a_{2}\right)+1$ ). What is the smallest value of $k$ such that $a_{2016}$ is a multiple of $3^{4031} \cdot 7^{4027}$?
(A) 1271
(B) 1322
(C) 1331
(D) 1702
(E) 2048
6. The answer is (B). First, let's prove that $a_{n}=(k+1)^{n-1}$: we have $a_{1}=1=(k+1)^{1-1}$ and $a_{n+1}=k\left(a_{n}+\ldots+a_{1}\right)+1=k a_{n}+k\left(a_{n-1}+\ldots+a_{1}\right)+1=k a_{n}+a_{n}=(k+1) a_{n}$; by induction, we obtain the desired formula.

Now consider $a_{2016}=(k+1)^{2015}$: it is divisible by $3^{4031}=3^{2 \cdot 2015+1}$ if and only if $3^{3}$ divides $k+1$; similarly, $a_{2016}$ is divisible by $7^{4027}=7^{2 \cdot 2015-3}$ if and only if $k+1$ is divisible by $7^{2}$. In conclusion, $k=3^{3} \cdot 7^{2}-1=1322$ is the minimum admissible value.