Prove that $\lim_{n\to\infty}a_n=\frac{\sum_{i=1}^k2ia_i}{k(k+1)}$ 
Given that a sequence $(a_n)$ satisfies $a_{n+k} = \dfrac{a_n + a_{n+1} + \cdots + a_{n+k-1}}{k}$ for $n\geq 1,$ where $k\in\mathbb{N},$ prove that $\lim\limits_{n\to\infty} a_n = \dfrac{2a_1}{k(k+1)}+\dfrac{4a_2}{k(k+1)}+\cdots +\dfrac{(2k)a_k}{k(k+1)}.$

I am not sure how to go about doing this. However, I understand why this might be the case. Informally, the $j$th term, where $1\leq j\leq k$ eventually occurs $j$ times more often than the first term. Also, when the first $k$ terms are $1,$ all the terms in the sequence are one, as evidenced by the proof below.
Proof: We proceed by strong induction. Suppose the first $k$ terms are $1.$ Then the $(k+1)$th term is also $1$ (since it is the average of the first $k$ terms). Assume that the first $k+m$ terms are $1$ for some $m\in\mathbb{N}.$ Then we have $a_{m+1} = \cdots = a_{k+m} = 1$ and $a_{k+m+1} = \dfrac{a_{m+1} + \cdots + a_{k+m}}k = 1,$ so the first $k+m+1$ terms are $1.$ Hence by strong induction, all the terms of the sequence are $1$ if the first $k$ terms are $1.$
Since the $j$th term, $1\leq j\leq k$ occurs $j$ times as often as the first term, we have that $\dfrac{1}a + \dfrac{2}a + \cdots + \dfrac{k}a = 1,$ where $\dfrac{1}a$ is the coefficient of $a_1$ in the limit of the sequence. Hence $\dfrac{k(k+1)}{2a} = 1\Rightarrow a = \dfrac{k(k+1)}2\Rightarrow \lim\limits_{n\to\infty} a_n = \dfrac{a_1}a + \dfrac{2a_2}a +\cdots + \dfrac{ka_k}a = \dfrac{2a_1}{k(k+1)} + \cdots + \dfrac{(2k)a_k}{k(k+1)}.$ 
Obviously the above reasoning is too informal to be considered a proof, so I am not sure how to prove this. If I could prove that the $j$th term eventually occurs $j$ times as often as the first term in the limit, it would be enough to prove this.

Based on the comment by Paramanand Singh, we see that we have the relation
$kx_{n+k} = x_{n+k-1} + x_{n+k-2} + \cdots + x_n.$
Hence $kx_{n+k}+(k-1)x_{n+k-1} +\cdots + 2x_{n+2} +x_{n+1} = kx_{n+k-1}+(k-1)x_{n+k-2}+\cdots + 2x_{n+1}+x_n.$
Repeating this process $n-1$ more times, we see that $kx_{n+k}+(k-1)x_{n+k-1}+\cdots + 2x_{n+2}+x_{n+1}=kx_k+(k-1)x_{k-1}+\cdots + 2x_2+x_1\,\forall n\geq 0\tag{1}$ Hence we have that \begin{align*}\lim\limits_{n\to\infty} kx_{n+k} + (k-1)x_{n+k-1} + \cdots + x_n &= (k+(k-1)+\cdots + 1)\lim\limits_{n\to\infty} x_n \\
&= \dfrac{k(k+1)}2 \lim\limits_{n\to\infty} x_n\tag{2}\end{align*}
as $\lim\limits_{n\to\infty} x_{n+j} = \lim\limits_{n\to\infty} x_n$ for $0\leq j\leq k$ (observe that since the limit exists, any subsequence converges to that limit, so it doesn't matter whether $k$ is finite).
However, we also have from $(1)$ that $\lim\limits_{n\to\infty} kx_{n+k} + (k-1)x_{n+k-1} + \cdots + x_n = kx_k + (k-1)x_{k-1}+\cdots + x_1,$ so $(2)$ gives that $\lim\limits_{n\to\infty} x_n = \dfrac{\sum_{j=1}^k 2jx_j}{k(k+1)},$ as required.
