Find all functions $f(f(f(...(f(x_1,x_2),x_3),...),x_{2016}))=x_1+x_2+...+x_{2016}$ I am trying to solve the functional equation:

Find all functions $f:\mathbb R^2\rightarrow \mathbb R$ such that for all $\left \{x_1,x_2,...,x_{2016} \right \}\subset \mathbb R$:
  $$f(f(f(...(f(x_1,x_2),x_3),...),x_{2016}))=x_1+x_2+...+x_{2016}$$

My work so far:
$f(x,y)=f(0+0+..+0+x,y)$
$f(f(f(...(f(0,0),0),...),x),y)=0+0+...+0+x+y=x+y$
I need help here.
 A: Considering the outer most iteration (the one with $x_{2016}$), we obviously have $f(x,y) = g(x)+y$ for some function $g$, so:
$\begin{align*}x_1+x_2+\ldots+x_{2015}+x_{2016}&=  f(f(f(\ldots f(x_1,x_2),\ldots,x_{2016}) \\
&= g(g(\ldots g(x_1)+x_2 \ldots )+x_{2015})+x_{2016} \\
\implies x_1+x_2+\ldots +x_{2015}\qquad \quad\,&=g(g(\ldots g(x_1)+x_2 \ldots )+x_{2015})  \end{align*}$
Setting $x_1 = \dots = x_{2014} =0$ and $c:=g(\ldots g(x_1)+x_2 \ldots ) = g(\ldots g(0)+0 \ldots )$ we get
$y = g(c+y)\;\forall y$ 
We subsitute $y$ with $t-c$  and get
$t-c = g(t)\;\forall t \qquad (*)$
Pluggin this equation back in the original results in
$\begin{align*}x_1+x_2+\ldots+x_{2016} &=  f(f(f(\ldots f(x_1,x_2),\ldots,x_{2016}) \\ &\overset{(*)}{=} (x_1-c)+(x_2-c)+\ldots+(x_{2015}-c)+x_{2016} \\ &= x_1+x_2+\ldots +x_{2016} - 2015c\end{align*}$
Therefore $c=0$ and so $g(t) = t$. And this implies $f(x,y)=x+y \;\forall x,y$.
A: So lets assume $f$ is a real function an
$$f(f(f(\ldots f(f(x_1,x_2),x_3)\ldots),x_{2016})= x_1+x_2+\ldots + x_{2016} \tag{1}$$
We now apply 
$$f(.,x_{2017}) \tag{2}$$
on the LHS and RHS of $(1)$ and we get
$$f( f(f(f(\ldots f(f(x_1,x_2),x_3)\ldots),x_{2016}) ,x_{2017}) = f(x_1+x_2+\ldots + x_{2016},x_{2017}) \tag{3}$$
If we change the variable names in $(1)$ , $x_{i}$ by $x_{i+1}$, we get
$$f(f(f(\ldots f(f(x_2,x_3),x_4)\ldots),x_{2017})= x_2+x_3+\ldots + x_{2017} \tag{4}$$
In $(4)$ now we substitute $x_2$ by $f(x_1,x_2)$ and get
$$f(f(f(\ldots f(f( f(x_1,x_2) ,x_3),x_4)\ldots),x_{2017})= f(x_1,x_2) +x_3+\ldots + x_{2017} \tag{5}$$
The LHS of $(3)$ and $(5)$ are the same. So the RHS of $(3)$ and $(5)$ must be equal and we get
$$f(x_1+x_2+\ldots + x_{2016},x_{2017}) = f(x_1,x_2) +x_3+\ldots + x_{2017} \tag{6}$$
Here  we set  $x_1=0$, $x_2=0$, ..., $x_{2015}=0$ and get
$$f(x_{2016},x_{2017})=f(0,0)+x_{2016}+x_{2017} \tag{7}$$
We can use $(7)$ to evaluate the nested function expression of the LHS of $(1)$. You can start with the outer or with the inner expression. Finally you will get
$$f( f(f(f(\ldots f(f(x_1,x_2),x_3)\ldots),x_{2016}) ,x_{2017})= 2015 f(0,0)+x_1+x_2+x_3+\cdots+x_{2016} \tag{8}$$
The RHS of $(1)$ and $(8)$ are equal, so ist must be
$$2015f(0,0)=0 \tag{9}$$
From $(9)$ and $(7)$ we see that
$$f(x,y)=x+y \tag{10}$$
as expected.
