<s>[INST] Is $f(n)=\sqrt{n}$ the only function from $\Bbb{N}_0$ to $[0,\infty)$, with $f(100)=10$ and $\sum_{k=0}^n\frac{1}{f(k)+f(k+1)}=f(n+1)$? 
Let $f:\mathbb{N_0} \rightarrow \left[0,\infty\right)$ be a function such that
a) $f(100)=10$
b)$\dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)}=f(n+1)$ for all $n\geq 0$.
Evaluate $f(n)$.

Clearly, $f(n)=\sqrt{n}$, satisfies the conditions, but my question is, is it the only function possible?
I don't see how using the value at $n=100$ would prove the function to be $\sqrt{n}$, however if instead of $100$, $f(1)=1$ was given, then we can prove by induction that $\sqrt{n}$ would be the only function by starting from the first term and finding the next term.
 [/INST] For $n\geq 1$,
\begin{align}
f(n+1) &= \dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)}\\
& = f(n)+\dfrac{1}{f(n)+f(n+1)}\\
\implies f(n+1)^2&=1+f(n)^2 =n+f(1)^2\\
\implies f(n) &= \sqrt{n-1+f(1)^2}.
\end{align}
Further, we have $$10 = f(100)=\sqrt{99+f(1)^2}\implies f(1)=1.$$
Finally, we also derive $$f(1)=\dfrac{1}{f(0)+f(1)}\implies f(0)=0.$$
Hence, we conclude that $f(n)=\sqrt{n}.$
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