5. The sequence $a_{n}$ is defined as follows:

$a_{1}=2, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{999}$.
Answer: 999000.

Solution: Consider the beginning of the sequence: $a_{1}=2, a_{2}=6, a_{3}=12, a_{4}=20, \ldots$

We can notice a pattern - the difference between consecutive terms forms an arithmetic progression: $4,6,8, \ldots$.

From this, we get the formula for the $\mathrm{n}$-th term $a_{n}=n \cdot(n+1)$, which can be proven by induction.

Thus, $a_{999}=999 \cdot 1000=999000$.