Solving $\sqrt[2k+1]{x+1}+\sqrt[2k+1]{x+2}+\sqrt[2k+1]{x+3}+...+\sqrt[2k+1]{x+n}=0$ I noticed that when graphing the function $f(x)=\sqrt[2k+1]{x+1}+\sqrt[2k+1]{x+2}+\sqrt[2k+1]{x+3}+...+\sqrt[2k+1]{x+n}$, its roots seemed to behave predictably, that is, $x=-\frac{n+1}{2}$ seems to always be the only real zero of $f(x)$. However, I'm not quite sure how to prove this. There is no way for me to expand this into a polynomial with integer roots so instead I tried to give it a bound using Am-Gm.
This gives, $f(x)≥n\big((x+1)(x+2)...(x+n)\big)^{\frac{1}{2k+1}}$. Setting $f(x)=0$ gives $\big((x+1)(x+2)...(x+n)\big)^{\frac{1}{2k+1}}≤0$ which implies that $(x+1)(x+2)...(x+n)≤0$ since $2k+1$ is odd and preserves the signs. However, I can't deduce anything from this result.
I also found $f'(x)=\frac{1}{2k+1}\Big((x+1)^{\frac{-2k}{2k+1}}+(x+2)^{\frac{-2k}{2k+1}}+...+(x+n)^{\frac{-2k}{2k+1}}\Big)$  and 
$f''(x)=\frac{-2k}{(2k+1)^2}\Big((x+1)^{-\frac{4k+1}{2k+1}}+(x+2)^{-\frac{4k+1}{2k+1}}+...+(x+n)^{-\frac{4k+1}{2k+1}}\Big)$. However, nothing significant jumps out about these two results that I could use.
So could someone help identify a way to "naturally" find the root, namely, not an educated guess that seems to work, and prove that it is the only real root which is what desmos' graph suggests?

Since $g(x)=\sqrt[2k+1]{x}$ is a strictly increasing function, therefore $f(x)$ is  also a strictly increasing function. This means that, if there exist a real solution, then there is only one real solution.
We know that, the function $g(x)$ is odd. Using this property and setting $x+1=-(x+n)$ leads to:
$$\begin{align}&x+1=-(x+n)\\
&x+2=-(x+(n-1))\\
&x+3=-(x+(n-2))\\
&\cdots\cdots\cdots\\\
&x+n=-(x+1)\end{align}$$
This implies,
$$\begin{align}&\sqrt [2k+1]{x+1}=-\sqrt [2k+1]{(x+n)}\\
&\sqrt [2k+1]{x+2}=-\sqrt [2k+1]{(x+(n-1))}\\
&\sqrt [2k+1]{x+3}=-\sqrt [2k+1]{(x+(n-2))}\\
&~~~\cdots\cdots\cdots\\\
&\sqrt [2k+1]{x+n}=-\sqrt [2k+1]{(x+1)}\end{align}$$
Summing up the equations side-by-side, we have:
$$f(x)=-f(x)\implies f(x)=0.$$
Therefore, the only real solution is $$\begin{align}&x+1=-(x+n)\\
\implies &x=-\frac {n+1}{2}.\end{align}$$
