A sequence $a_1, a_2, a_3 ...$ satisfies $$a_{n+3} = 2a_{n+2} - 2a_{n+1} + a_n$$ for $n= 1,2,3...$ If $a_1 = 11, a_2 = 222$ and $a_3 = 7777$, find $a_{2016}$.
So the first thing I did was to find $a_4$ by plugging in the given numbers which give $a_4 = 15121$, and I computed a few more terms after that, but I figured that at this rate it would take me a really long time to figure out any patterns (if at all).
Then I tried to mainipulate the given equation $a_{n+3} = 2a_{n+2} - 2a_{n+1} + a_n$ so that it looks like $$a_{n+3} - a_{n+2} = (a_{n+2} - a_{n+1}) - (a_{n+1} - a_n). \space \space \space —(1)$$ Then letting $b_n = a_{n+1} - a_n$, I can equate $(1)$ to $$b_{n+2} = b_{n+1} - b_n.\space \space \space —(2)$$
However, I wasn't able to figure out anything much afterwards, and any help would really be appreciated!
Hint: $a_7 = 11, a_8 = 222, a_9 = 7777$.
Here is a slightly more sophisticated answer: The characteristic polynomial of the recurrence is $x^3-(2x^2-2x+1)=(x - 1) (x^2 - x + 1) = \Phi_1(x) \Phi_6(x)$, where $\Phi_n$ is the $n$-th cyclotomic polynomial. Therefore, its companion matrix $A$ satisfies $A^6=I$ because $x^6-1=\Phi_1(x) \Phi_2(x)\Phi_3(x)\Phi_6(x)$ is a multiple of the characteristic polynomial. As a consequence, the sequence is periodic with period $6$.