Generalizations of $\sum_{m=3n+2}^{\infty}\phi^m=\phi^{3n}$ and $\sum_{m=13n+1}^{\infty}(\sqrt2-1)^m=\dfrac{(\sqrt2-1)^{13n}}{\sqrt2}$ I noticed that the following identies hold with the help of wolfram alpha and oeis. I'm sure they're well-known, but I'd like to know how they generalize.
$$\sum_{m=3n+2}^{\infty}\phi^m=\phi^{3n}=(-1)^n(a_n-b_n\sqrt5),$$
where $\phi=\dfrac{\sqrt5-1}2,a_0=1,b_0=0,$ and $a_n$ and $b_n$ are the numerator and denominator respectively of the $n$th convergent of the continued fraction $\sqrt5=2+\dfrac1{4+\dfrac1{4+\dfrac1{4+\cdots}}}$
Similarly,
$$\sum_{m=13n+1}^{\infty}(\sqrt2-1)^m=\dfrac{(\sqrt2-1)^{13n}}{\sqrt2}=(-1)^n\dfrac{c_{13n}-d_{13n}\sqrt2}{\sqrt2},$$
where $c_0=1,d_0=0,$ and $c_n$ and $d_n$ are the numerator and denominator respectively of the $n$th convergent of the continued fraction $\sqrt2=1+\dfrac1{2+\dfrac1{2+\dfrac1{2+\cdots}}}$
Do similar identities exist for other values of $\dfrac{\sqrt{k^2+4}-k}2$ for positive integer $k$ (the previous examples are the cases $k=1,2$ respectively)?

This is a really awesome observation of yours. The main thing that will help you here is to think in purely algebraic terms. Notably, any product, sum, quotient, etc. involving only rational numbers and $\sqrt{k}$ can be written in the form $a+b\sqrt{k}$. In particular, if we define $\phi_k=\frac{\sqrt{k^2+4}-k}2$, notice that this is a root of
$$\phi_k^2+k\phi_k-1=0.$$
Next, define the function
$$f_k(n)=\sum_{m=n}^{\infty}\phi_k^m=\frac{\phi_k^m}{1-\phi_k}$$
where the latter expression is a well-known identity for geometric sums. We want to find two rational sequence $a_n$ and $b_n$ such that
$$f_k(n)=a_n+b_n\phi_k$$ 
This would be a very long post if we went through all the algebra to derive all the equations I'm gonna use (I started to write such a post) - so I won't where the calculation is routine. From the quadratic equation above, it follows that
$$f_k(0)=\frac{1}{1-\phi_k}=\frac{k+1}k+\frac{1}k\phi_k$$
which can be taken to mean that $a_0=\frac{k+1}{k}$ and $b_0=\frac{1}k$. Further, noting that $f_k(n+1)=\phi_kf_k(n)$, we write
$$a_{n+1}+b_{n+1}\phi_k=a_n\phi_k + b_n\phi_k^2$$
and, after rearranging terms and seeing that $\phi_k^2=1-k\phi_k$, we get
$$a_{n+1}+b_{n+1}\phi_k=b_n+(a_n-kb_n)\phi_k$$
implying that $a$ and $b$ are governed by the following recurrence
$$\begin{array}{ll}a_0=\frac{k+1}{k} && b_0=\frac{1}k \\ a_{n+1}=b_n && b_{n+1}=-kb_n+a_n \end{array}$$
wherefrom we can easily eliminate $a$ to get the equations
$$b_0=\frac{1}k$$
$$b_1=\frac{1}k$$
$$b_{n+2}'=-kb_{n+1}'+b_n'$$
And then we have $f_k(n)=b_{n-1}+b_n\phi_k$, which is a pretty interesting relation in and of itself, especially given how $b$ looks kind of like the Fibonacci sequence. Actually, if $k=1$, then, after $3$ places, the sequence $b$ is the Fibonacci sequence, just with its signs alternating. However, to further elucidate the connection, set $b_n'=k(-1)^nb_n$. Then we have
$$b_0'=1$$
$$b_1'=-1$$
$$b_{n+2}'=kb_{n+1}'+b_n'$$
which, for $k=1$ is a shift of the Fibonacci sequence. Also, we have $f_k(n)=(-1)^n(b_n'\phi_k-b_{n-1}')$
Better yet, this series very directly relates to the series of convergents for $\phi_k$. Another property of $\phi_k$ following from the quadratic it solves is that
$$\phi_k=\frac{1}{k+\phi_k}$$
meaning that the continued fraction for $\phi_k$ is $[0;k,k,\ldots]$ and, therefore, that the series of convergents $x$ can be defined by
$$x_0=0$$
$$x_{n+1}=\frac{1}{k+x_n}$$
and that, if we substitute $x_n=\frac{p_n}{q_n}$ for coprime sequences of integers $p$ and $q$, we get
$$\frac{p_n}{q_n}=\frac{0}{1}$$
$$\frac{p_{n+1}}{q_{n+1}}=\frac{1}{k+\frac{p_n}{q_n}}=\frac{q_n}{kq_n+p_n}$$
which, again, yields a nice recurrence relation
$$\begin{array}{ll}p_0=0 &&q_0=1\\p_{n+1}=q_n && q_{n+1}=kq_n+p_n\end{array}$$
which looks fairly familiar, especially when we eliminate $p$ from the equation and get
$$q_0=1$$
$$q_1=k$$
$$q_{n+2}=kq_{n+1}+q_n$$
where the convergents are just $x_n=\frac{q_{n-1}}{q_n}$ - but, hey, isn't that the same recurrence relation as defines $b'$, just with different boundary conditions? (hint: yes)
From here, note that, for $k=1$, then $q_n=(-1)^{n+1} b_{n+3}$ and $p_n=(-1)^n a_{n+3}$, so that means that $f_1(n+3)=(-1)^n(p_n-q_n\phi_1)$ where $p_n/q_n$ are the convergents of $\phi_1$. Presumably, the exact relation you cite, using $\sqrt{5}$ is just a stone's throw from there. Unfortunately, far as I can tell, there is no shift of $b$ that would align it with $q$ for any $k>1$, but perhaps using this groundwork, where you have two similar recurrence relations you can find something. (If I think of something, I'll update this answer; I suspect that, given the relation of $a_n$ and $b_n$, we could easily find $c_n$ and $d_n$ such that $a_n+b_n\phi_k=c_n+d_n\phi_k'$ for some other irrational $\phi_k'$ satisfying certain properties and there might be some trick therein to derive an identity; I would tend to believe that something more exists)
