<s>[INST] 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.
 [/INST] 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.
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