Year in a sequence (4)

Relevant wiki: Linear Recurrence Relations

From the first few terms of $a_n$ , it seems we can claim that $\displaystyle \sum_{k=1}^n a_k = 4a_{n-1}$ . Let prove by induction that the claim is true for all $n \in \mathbb N$ .

Proof:

For $n=2$ , we have $\displaystyle \sum_{k=1}^2 a_k = a_1+a_2 = a_1 + \dfrac {2(2+1)}2 a_1 = 4a_1$ Therefore, the claim is true for $n=2$ .

Assuming that the claim is true for $n$ , then we have:

$\begin{aligned} \sum_{k=1}^{\color{#D61F06}n+1} a_k & = \sum_{k=1}^n a_k + a_{n+1} \\ & = 4a_{n-1} + \frac {2(n+2)}{n+1}a_n \\ & = \frac {4n}{2(n+1)}a_n + \frac {2(n+2)}{n+1}a_n \\ & = \frac {4n + 4n+8}{2(n+1)}a_n \\ & = 4a_{{\color{#D61F06}(n+1)}-1} \end{aligned}$

Therefore, the claim is true for $n+1$ and hence true for all $n \in \mathbb N$ . $\square$

$\implies \dfrac {a_{2016}}{a_1+a_2+a_3+\cdots + a_{2017}} = \dfrac {a_{2016}}{\sum_{k=1}^{2017} a_k} = \dfrac {a_{2016}}{4a_{2016}} = \boxed{\dfrac 14}$