Limit of a 31st order linear recurrence We have the recursive sequence $y_k=k$ for $1\leq k \leq 31$ and $31y_{k+1}=\sum_{i=0}^{30}y_{k-i}$ for $k\geq 31$.
How do we find the limit? 
This was an exam question and the answer given to me is $$\lim _{k\to \infty}y_k=\dfrac{1^2+2^2+3^2\dots+31^2}{1+2+3+\dots+31}=21.$$
I fail to understand how this makes any sense. Also, it would be nice if answers are elementary as this was on a mock test for JEE Main.
 A: $$
y_{k+1}=\frac{1}{31}\sum_{i=0}^{30} y_{k-i}
$$
Consider that the characteristic polynomial for this recurrence relation has a root at 1. Dividing by $(x-1)$, we can get a new characteristic polynomial. Let's assume that
$$
\sum_{n=0}^{30} (n+1)y_{k+n}=c_k
$$
Now, $c_k=c$ is a constant. Proof? Consider the difference between two successive values, here:
$$
\sum_{n=1}^{31} ny_{k+n}-\sum_{n=0}^{30} (n+1)y_{k+n}=c_{k+1}-c_k\\
31y_{k+31}+\sum_{n=1}^{30}ny_{k+n}-\sum_{n=1}^{30} ny_{k+n}-\sum_{n=0}^{30}y_{k+n}=c_{k+1}-c_k\\
31y_{k+31}=c_{k+1}-c_k+\sum_{n=0}^{30}y_{k+n}
$$
Which, from our original recurrence relation, means that $c_{k+1}=c_k$, and therefore it is a constant.
So what is $c$? Let's look at the defining values - $y_k=k$ for $1\leq k\leq 31$. Take our new relation for $k=1$. So
$$
c = \sum_{n=0}^{30} (n+1)(n+1) = \sum_{n=0}^{30} (n+1)^2
$$
Now, in the limit, the values will all be equal - that is, $\lim_{n\to \infty} y_k = \lim_{n\to \infty} y_{k+1} = y$. So we have
$$
c = \sum_{n=0}^{30} (n+1)y\\
y = \frac{c}{\sum_{n=0}^{30} (n+1)} = \frac{\sum_{n=0}^{30} (n+1)^2}{\sum_{n=0}^{30} (n+1)}
$$
