A sequence to achieve $\frac{1}{a_{2016}}$ It is given that $a_ka_{k-1} + a_{k-1}​a_{k-2}​ = 2a_k a_{k-2}$  , $k\geq3$ and $a_1=1$.
We have $S_q= \sum_{k=1}^{q} \frac{1}{a_k} $ and given that $\frac{S_{2q}}{S_{q}}$ is independent of q then $\frac{1}{a_{2016}}$ is = ? 
I think no information is given to find $a_2$. How should we approach this problem?
 A: You get some information, I just don't think you get enough (unless I'm misreading the problem).
Given $a_1 = 1, a_2$, we can determine $a_3 = \frac{a_2}{2-a_2}$ and $a_4 = \frac{a_2^2}{3a_2-2a_2^2}$.  Based on the conditions imposed on $S_q$, we also have
$$
\frac{S_2}{S_1} = \frac{S_4}{S_2}
$$
$$
1+\frac{1}{a_2} = \frac{1+\frac{1}{a_2}+\frac{2-a_2}{a_2}+\frac{3a_2-2a_2^2}{a_2^2}}{1+\frac{1}{a_2}}
$$
which can be simplified to
$$
\frac{1+a_2}{a_2} = \frac{6-2a_2}{1+a_2}
$$
or
$$
1-4a_2+3a_2^2 = 0
$$
which yields the two solutions $a_2 = 1, a_2 = \frac{1}{3}$.  These two solutions yield the two series $1, 1, 1, 1, \ldots$ and $1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \ldots$, and hence two different values of $\frac{1}{a_{2016}}$, so I'm not sure how one should proceed.
A: You can find $a_2$ by solving three simultaneous equations ($a_1$ fixed at $1$):
$$
 a_3 a_2 + a_2 = 2_a3 \implies a_2 = \frac{2a_3}{a_3+1} \\
a_4a_3 + a_3a_2 = 2a_4a_2\implies a_4a_3 + a_3\frac{2a_3}{a_3+1} = 2a_4\frac{2a_3}{a_3+1}\implies a_4 = \frac{2a_3}{3-a_3} \\
\frac{S_4}{S_2} = \frac{S_2}{S_1} \implies \frac{1+\frac{a_3+1}{2a_3}+\frac{1}{a_3}+\frac{3-a_3}{2a_3}}{1+\frac{a_3+1}{2a_3}} = 1+\frac{a_3+1}{2a_3}
$$
That last equation simplifies to $$5a_3^2-6a_3+1=0$$.  This has two solutions.
If we take $a_3 = 1$ then $a_2 = 1$ and $a_i = 1$ for all $i$, and in that case 
$$
\frac{1}{a_{2016}} = 1$$
However, the answer your problem poser intended was probably the solution taking $a_3=\frac{1}{5}$.  Then $a_2 = \frac{1}{3}$ and it is easy to show that in general 
$$a_i = \frac{1}{2i-1}$$
Thus the answer you are probably looking for is 
\frac{1}{a_{2016}} = 2015$$
