5. The function $\mathbf{y}=f(x)$ satisfies the following conditions:

a) $f(4)=2$;

b) $f(n+1)=\frac{1}{f(0)+f(1)}+\frac{1}{f(1)+f(2)}+\ldots+\frac{1}{f(n)+f(n+1)}, n \geq 0$.

## Find the value of $f(2022)$.
Solution. For any $\quad n \geq 0 \quad f(n+2)-f(n+1)=\frac{1}{f(n+2)+f(n+1)}, \quad$ from which $f^{2}(n+2)=f^{2}(n+1)+1$. Using the condition $f(4)=2$, we get:

$$
\begin{gathered}
4=f^{2}(4)=f^{2}(3)+1 \Rightarrow f(3)=\sqrt{3} \\
3=f^{2}(3)=f^{2}(2)+1 \Rightarrow f(2)=\sqrt{2} \\
2=f^{2}(2)=f^{2}(1)+1 \Rightarrow f(1)=1 \\
1=f(1)=\frac{1}{f(0)+f(1)}=\frac{1}{f(0)+1} \Rightarrow f(0)=0
\end{gathered}
$$

Using mathematical induction, we will show that $f(n)=\overline{\text {. }}$. The base case of the induction has been proven above. For the inductive step, we use the relation $f^{2}(n+2)=f^{2}(n+1)+1:$

$$
f^{2}(n+2)=f^{2}(n+1)+1=(n+1)+1=n+2 \Rightarrow f(n+2)=\sqrt{n+2}
$$

It is not difficult to show that the function indeed satisfies the condition of the problem: if,

$$
\frac{1}{\sqrt{0}+\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1} \text { then }
$$

$$
\frac{1}{\sqrt{0}+\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}+\frac{1}{\sqrt{n+1}+\sqrt{n+2}}=\sqrt{n+1}+\frac{1}{\sqrt{n+1}+\sqrt{n+2}}=\frac{n+2+\sqrt{(n+1)(n+2)}}{\sqrt{n+1}+\sqrt{n+2}}=\sqrt{n+2}
$$

Thus, $f(2022)=\cdot \overline{2 \pi 2 \overline{2}}$

![](https://cdn.mathpix.com/cropped/2024_05_06_de9e429f583f4f49f4cdg-3.jpg?height=57&width=272&top_left_y=891&top_left_x=298)