SOLUTION. First, let's calculate the values of the expression $V(n)$ for several natural numbers $n \geq 3:$

| $n$ | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $V(n)$ | $5 \frac{1}{3}$ | $5 \frac{1}{3}$ | $6 \frac{2}{7}$ | $7 \frac{2}{3}$ | $10 \frac{2}{3}$ | 2 | $7 \frac{5}{9}$ | $9 \frac{2}{9}$ | $10 \frac{14}{29}$ | $11 \frac{13}{21}$ | $12 \frac{40}{57}$ | $13 \frac{28}{37}$ |

From the table, we see that $V(n) \geq 2$ for all $n \in\{3,4, \ldots, 14\}$, with $V(8)=2$. We will show that for all $n \geq 9$, it holds that $V(n)>2$.

By successive manipulation of the expression $V(n)-2$, we get (we know that $V(8)-2=0$)

\[
\begin{aligned}
V(n)-2 & =\frac{n^{3}-10 n^{2}+17 n-4}{n^{2}-10 n+18}-2=\frac{n^{3}-12 n^{2}+37 n-40}{n^{2}-10 n+18}= \\
& =\frac{(n-8)\left(n^{2}-4 n+5\right)}{(n-5)^{2}-7}=\frac{(n-8)\left((n-2)^{2}+1\right)}{(n-5)^{2}-7} .
\end{aligned}
\]

For $n \geq 9$, both the numerator and the denominator of the last fraction are positive numbers. Therefore, $V(n)-2>0$ for every $n \geq 9$.

Answer: The smallest value of the fraction $V(n)$ for all natural numbers $n>2$ is 2; this value is attained by the expression $V(n)$ for $n=8$.
## 4. Find the smallest value of the fraction

$$
V(n)=\frac{n^{3}-10 n^{2}+17 n-4}{n^{2}-10 n+18}
$$

$where\ n$ is any natural number greater than 2.