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.
 A: (Fill in the gaps as needed. If you're stuck, show your work and explain why you're stuck.)
Ignore condition (a) for now (This just gives us a specific solution).
Claim: Prove that condition $b$ is equivalent to

*

*$ \frac{1}{ f(0) + f(1) } = f(1)$, and

*$ f(n)^2 - f(n-1)^2 = 1$ for $n\geq 2$
Corollary:  The family of solutions that satisfies $(b)$ is given by
$$ \begin{align} f(0) & = a, \\ 
f(1) & = \frac{1}{2} ( \sqrt{ a^2 + 4} - a ), \\
f(n) & = \sqrt{ f(1)^2 + n-1 } \end{align}. $$
Corollary: Hence, using condition $a$, since $f(100) = 10$, so $f(1) = 1$ and $ f(0) = a = 0$. Thus, the solution is uniquely $ \sqrt{n}$.

Notes

*

*If we just care about condition $b$, then for $ n \geq 1$,  $\sqrt{n} \geq f(n) > \sqrt{n-1}$. You can show that by showing that the range of $ \frac{1}{2} \sqrt{a^2 + 4} - a ) $ on $[0, \infty)$ is $ (0, 1]$.

A: Using (b) with $n$ and then (b) with $n-1$ and subtracting, we get
\begin{align*}
f(n+1) - f(n) &= \bigg( \dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)} \bigg) \\
&\qquad{}- \bigg( \dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n-1)+f(n)} \bigg) \\
&= \frac1{f(n)+f(n+1)}.
\end{align*}
Therefore $f(n+1)^2 - f(n)^2 = 1$ for all $n\ge0$, which immediately fixes all the values given any single value.
A: 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 $f(n)=\sqrt{n}.$
