$\underline{\text { Glibov A. }}$.

Let $n$ be a natural number. We call a sequence $a_{1}, a_{2}, \ldots, a_{n}$ interesting if for each $i=1,2$, $\ldots, n$ one of the equalities $a_{i}=i$ or $a_{i}=i+1$ holds. We call an interesting sequence even if the sum of its terms is even, and odd otherwise. For each odd interesting sequence, we found the product of its terms and wrote it on the first sheet. For each even sequence, we did the same and wrote it on the second sheet. On which sheet is the sum of the numbers greater and by how much? (Give the answer in terms of $n$.)
Denoting the sum containing the term $2 \cdot 3 \cdot \ldots \cdot n(n+1)$ by $A_{n}$, and the other by $B_{n}$, we will prove the equality $A_{n}-B_{n}=1$ by induction.

Base case. $A_{1}-B_{1}=2-1=1$.

Inductive step. Represent the sum $A_{n}$ as $A^{\prime}+A^{\prime \prime}$, where $A^{\prime}$ contains all terms of the form $a_{1} a_{2} \ldots a_{n-1}(n+1)$, and $A^{\prime \prime}$ contains all terms of the form $a_{1} a_{2} \ldots a_{n-1} n$. Since the sum $A^{\prime}$ contains the term $2 \cdot 3 \cdot \ldots \cdot n(n+1)$, for each term $a_{1} a_{2} \ldots a_{n-1}(n+1)$ in it, the sequence $a_{1}, a_{2}, \ldots, a_{n-1}$ has the same parity as the sequence $2,3, \ldots, n$. Therefore, $A^{\prime}=(n+1) A_{n-1}$. Correspondingly, in the sum $A^{\prime \prime}$, for each term $a_{1} a_{2} \ldots a_{n-1}$, the parity of the sequence $a_{1}, a_{2}, \ldots, a_{n-1}$ is opposite to the parity of the sequence $2,3, \ldots, n$. Then $A^{\prime \prime}=n B_{n-1}$, from which $A_{n}=A^{\prime}+A^{\prime \prime}=(n+1) A_{n-1}+n B_{n-1}$. Similarly, $B_{n}=n A_{n-1}+(n+1) B_{n-1}$. Therefore, $A_{n}-B_{n}=(n+1) A_{n-1}+n B_{n-1}-n A_{n-1}-(n+1) B_{n-1}=A_{n-1}-B_{n-1}=1$.

## OTBet

When $n \equiv 0,1(\bmod 4)$, the sum on the second sheet is greater by 1, and for other $n$ - the opposite. In other words, the sum that contains the term $2 \cdot 3 \cdot \ldots \cdot n(n+1)$ is greater by 1.