The terms of a number sequence are:

$$
\begin{aligned}
& a_{0}=1, \quad a_{1}=1-\left(\frac{1}{2}\right)^{2} a_{0}, \quad a_{2}=1-\left(\frac{2}{3}\right)^{2} a_{1} \\
& a_{3}=1-\left(\frac{3}{4}\right)^{2} a_{2}, \ldots, \quad a_{n}=1-\left(\frac{n}{n+1}\right)^{2} a_{n-1}
\end{aligned}
$$

Write an expression for $a_{n}$ that does not use any previous terms of the sequence. Also, write such an expression for the following product:

$$
a_{0} a_{1} a_{2} a_{3} \ldots a_{n}
$$
I. The terms of the series with indices $1-3$ can be written as follows:

$$
\begin{aligned}
& a_{1}=1-\frac{1}{2^{2}}=\frac{1}{2^{2}}\left(2^{2}-1^{2}\right) \\
& a_{2}=1-\frac{1}{3^{2}}\left(2^{2}-1^{2}\right)=\frac{1}{3^{2}}\left(3^{2}-2^{2}+1^{2}\right) \\
& a_{3}=1-\frac{1}{4^{2}}\left(3^{2}-2^{2}+1^{2}\right)=\frac{1}{4^{2}}\left(4^{2}-3^{2}+2^{2}-1^{2}\right)
\end{aligned}
$$

Based on these, we conjecture the following expression for $a_{n}$:

$$
\begin{gathered}
a_{n}=\frac{1}{(n+1)^{2}}\left[(n+1)^{2}-n^{2}+(n-1)^{2}-(n-2)^{2}+\ldots+\right. \\
\left.+(-1)^{n} \cdot 2^{2}+(-1)^{n+1} \cdot 1^{2}\right]
\end{gathered}
$$

In the parentheses, the squares of the numbers are listed in reverse order, from $(n+1)^{2}$ to $1^{2}$, with every second one being a subtrahend. By pairing the consecutive terms within the brackets, each pair equals the sum of the bases, since the difference between the bases is always 1; the beginning of the expression thus becomes:

$$
[(n+1)+n]+[(n-1)+(n-2)]+\ldots
$$

and the end, for even or odd $n$, is:

$$
\ldots+(3+2)+1^{2}, \quad \text { or } \quad \ldots+(2+1)
$$

Thus, the expression is always the sum of the integers from $n+1$ to 1. Therefore,

$$
a_{n}=\frac{1}{(n+1)^{2}} \cdot \frac{(n+1)(n+2)}{2}=\frac{n+2}{2(n+1)}
$$

According to our conjecture, any term can be calculated from its index by dividing the quotient of the numbers 2 and 1 greater than the index by 2.

It is easy to verify that this matches the numbers derived from the specified form of the terms for $n=0,1,2,3$. We will show that if (3) holds for the index $n$, then it also holds for the next index $n+1$. Indeed, according to (1),

$$
a_{n+1}=1-\left(\frac{n+1}{n+2}\right)^{2} a_{n}=1-\frac{n+1}{2(n+2)}=\frac{n+3}{2(n+2)}
$$

which conforms to the stated rule of formation.

II. According to these, the product under (2), consisting of $n+1$ factors, has the product of the integers from 2 to $n+2$ in the numerator, and $2^{n+1}$ times the product of the integers from 1 to $n+1$ in the denominator. Thus, after simplification,

$$
a_{0} a_{1} a_{2} \ldots a_{n}=\frac{n+2}{2^{n+1}}
$$

Indeed, according to (4),

$$
a_{0} a_{1} a_{2} \ldots a_{n} \cdot a_{n+1}=\frac{n+2}{2^{n+1}} \cdot \frac{n+3}{2(n+2)}=\frac{n+3}{2^{n+2}}
$$

and this appears from $n+1$ in the same way as (5) does from $n$.

Lux Iván (Budapest, Rákóczi F. g. IV. o. t.)

Remark. Calculating the initial terms:

$$
a_{0}=1, \quad a_{1}=\frac{3}{4}, \quad a_{2}=\frac{2}{3}, \quad a_{3}=\frac{5}{8}, \quad a_{4}=\frac{3}{5}, \quad a_{5}=\frac{7}{12}
$$

The numerators and denominators of the consecutive terms show alternating growth and decrease. The denominators of the terms with even indices, which decrease, are odd. Writing these terms in the form:

$$
a_{0}=\frac{2}{2}, \quad a_{2}=\frac{4}{6}, \quad a_{4}=\frac{6}{10}
$$

by multiplying by 2, the regularity of (3) becomes apparent in the above sequence.