<s>[INST] 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?
 [/INST] 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 $$\begin{align}&x+1=-(x+n)\\
\implies &x=-\frac {n+1}{2}.\end{align}$$
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