4. If the product of three consecutive natural numbers $n-1, n$, and $n+1$ is increased by the middle number, the result is a number between 3000 and 4000. Determine these numbers.
4. First, we determine that $(n-1) n(n+1)+n=n^{3}$. We write the inequality $3000<n^{3}<4000$. We conclude that $\sqrt[3]{3} 000<n$ or $14.42<n$ and that $n^{3}<4000$ or $n<15.87$. Thus, $n=15$. The sought consecutive natural numbers are 14, 15, and 16.
Writing $(n-1) n(n+1)+n$ ..... 1 point
Simplifying the above expression to $=n^{3}$. . . ..... 1 point
Writing the inequality $3000<n^{3}<4000$ ..... 1 point
Reasoning: $14.42<n<15.87$. . . ..... 1 point
Solution $n=15$. . ..... 1 point
Answer: The sought numbers are $14,15,16$. ..... 1 point