Here's the question that I was instructed to do:
Write numbers 2, 2 on your paper. Write the sum of two numbers between them. We get 2, 4, 2. Repeat: 2, 6, 4, 6, 2. Repeat: 2, 8, 6, 10, 4, 10, 6, 8, 2. Find the sum of all numbers on the paper after repeating this 2017 times.
I tried using arithmetic and geometric progression, but to no avail, I can't find any sequence that would define them. (The sum is 4, 8, 20, 56, 164,... for the first, second, third and so on.) I have also tried using the quadratic formula.
Hint: you should be looking for a recurrence relation among the sums. If you have a middle number in the $n^{th}$ sum how many times does it contribute to the $(n+1)^{st}$ sum? How about the end $2$s?
Added much later: the end $2$s of one row are counted two times in the next, once at the outer $2$ and once in the sum one inboard. Every other entry in a row gets counted three times in the next row, once in its place and once in the sum on each side. If $a_n$ is the sum of the $n^{th}$ row, the recurrence is $a_{n+1}=3a_n-4$ because there are two $2$s that are only counted twice. By the usual technique for inhomogeneous linear recurrence relations, the solution is $a_n=b3^n+2$. If the first $2\ 2$ row is row $0$ this says $b=2$ and $a_n=2\cdot 3^n+2$, which matches your series.