The sequence $\mathrm{Az}\left(a_{i}\right)$ is defined as follows:

$$
a_{1}=1, \quad a_{n}=\frac{n+1}{n-1}\left(a_{1}+\cdots+a_{n-1}\right)
$$

Determine the value of $a_{1998}$.
I. solution. According to the formula, each term of the sequence is greater than the previous one, so the sequence is positive and monotonically increasing.

Multiply the $k$-th term by $(k-1)$, and the $(k-1)$-th term by $(k-2)$ for $(k>2$ integer):

\[
\begin{gathered}
(k-1) a_{k}=(k+1)\left(a_{1}+\cdots+a_{k-1}\right) \\
(k-2) a_{k-1}=k\left(a_{1}+\cdots+a_{k-2}\right)
\end{gathered}
\]

Add $k \cdot a_{k-1}$ to both sides here:

\[
(2 k-2) a_{k-1}=k\left(a_{1}+\cdots+a_{k-1}\right)
\]

We can divide (2) by (1) (since $a_{k}$ and $a_{1}+\cdots+a_{k-1}$ are both positive):

\[
2 \cdot \frac{a_{k-1}}{a_{k}}=\frac{k}{k+1}
\]

Writing (3) for $k=n, n-1, \ldots, 3$ and even for 2 (which is also true due to $a_{2}=3 a_{1}$), and then multiplying all of them, we get:

\[
2 \frac{a_{n-1}}{a_{n}} \cdot 2 \frac{a_{n-2}}{a_{n-1}} \cdot 2 \frac{a_{n-3}}{a_{n-2}} \cdots 2 \frac{a_{2}}{a_{3}} \cdot 2 \frac{a_{1}}{a_{2}}=\frac{n}{n+1} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n-1} \cdots \cdots \frac{3}{4} \cdot \frac{2}{3}
\]

which means

\[
2^{n-1} \frac{a_{1}}{a_{n}}=\frac{2}{n+1}
\]

so $a_{n}=(n+1) 2^{n-2}, a_{1998}=1999 \cdot 2^{1996}$.

II. solution. Calculating the first few elements of the sequence, $a_{1}=1, a_{2}=3, a_{3}=8, a_{4}=20$, it can be conjectured that $a_{n}=(n+1) 2^{n-2}$ for $n \geq 2$. Let's prove this by complete induction. We can calculate that it holds for $n=2,3,4$. Assume that for some integer $n(>4)$, $a_{n}=(n+1) 2^{n-2}$, and prove that it also holds for $(n+1)$, i.e., $a_{n+1}=(n+2) 2^{n-1}$. Since $a_{n}=\frac{n+1}{n-1}\left(a_{1}+\cdots+a_{n-1}\right)=(n+1) 2^{n-2}$, we have $a_{1}+\cdots+a_{n-1}=(n-1) \cdot 2^{n-2}$, so

\[
a_{n+1}=\frac{n+2}{n}\left(a_{1}+\cdots+a_{n}\right)=\frac{n+2}{n}\left((n-1) 2^{n-2}+(n+1) 2^{n-2}\right)=\frac{n+2}{n} \cdot 2 \cdot n \cdot 2^{n-2}=(n+2) 2^{n-1}
\]

which is what we wanted to prove.

Based on the solutions of Lajos Deli (Hajdúszoboszló, Hőgyes E. Gymnasium, 10th grade)