Determine the general term of the sequence $(a_n)_{n\ge1}$, strictly decreasing, of strictly positive numbers, which satisfies the properties:
a) $na_n \in \mathbb{N} \setminus \{0\}$ for every $n \in \mathbb{N} \setminus \{0\}$.
b) $\sum_{k=1}^{n} (-1)^{k-1} a_k \binom{n}{k} = \sum_{k=1}^{n} a_k$.
I tried taking $n=1,2,3,4$ to use induction but I couldn't find a pattern. Any ideas on how to approach it?
My approach: I'm pretty sure I made a lot of mistakes but this may help finding the correct proof.
Let's denote the general term of the sequence $(a_n)_{n\geq1}$ as $a_n = \frac{c}{n}$, where $c$ is some positive constant.
First, let's consider property a):
$$
na_n = n \cdot \frac{c}{n} = c.
$$
So, $na_n$ is always a positive integer, as required.
Now, let's consider property b). We need to find $c$ such that:
$$
\sum_{k=1}^{n} (-1)^{k-1} a_k \binom{n}{k} = \sum_{k=1}^{n} a_k.
$$
We substitute $a_k = \frac{c}{k}$ into the given equation:
$$
\sum_{k=1}^{n} (-1)^{k-1} \left(\frac{c}{k}\right) \binom{n}{k} = \sum_{k=1}^{n} \frac{c}{k}.
$$
For convenience, let's rewrite the binomial coefficient as:
$$
\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
Substituting this into our equation:
$$ \sum_{k=1}^{n} (-1)^{k-1} \left(\frac{c}{k}\right) \frac{n!}{k!(n-k)!} = \sum_{k=1}^{n} \frac{c}{k}.
$$
Rearranging the terms, we get:
$$
\begin{split}
c&\sum_{k=1}^{n} (-1)^{k-1} \frac{n!}{k!(n-k)!k} = c\sum_{k=1}^{n} \frac{1}{k} \\
c&\sum_{k=1}^{n} (-1)^{k-1} \frac{n!}{k!(n-k)!k} - c\sum_{k=1}^{n} \frac{1}{k} = 0 \\
c&\sum_{k=1}^{n} \frac{(-1)^{k-1}n!}{k!(n-k)!k} - c\sum_{k=1}^{n} \frac{1}{k} = 0
\end{split}
$$
Now, let's factor out $c$:
$$
c\left(\sum_{k=1}^{n} \frac{(-1)^{k-1}n!}{k!(n-k)!k} - \sum_{k=1}^{n} \frac{1}{k^2}\right) = 0 \iff
\sum_{k=1}^{n} \frac{(-1)^{k-1}n!}{k!(n-k)!k} = \sum_{k=1}^{n} \frac{1}{k}
$$
Given that this equation must hold for all $n$, we can conclude that the sum of the coefficients in both series must be equal.
$$
\begin{split}
\frac{(-1)^{k-1}n!}{k!(n-k)!k} &= \frac{1}{k} \\
(-1)^{k-1}n! &= \frac{(n-k)!k^2}{(n-k)!} \\
(-1)^{k-1}n! &= k^2
\end{split}
$$
Since this equation should hold for all $k$, the only way this can happen is if $n! = k^2$. This is not possible for all $n$ and $k$, so we need to revise our assumption about $a_n$.
The correct approach is to consider a sequence such that $a_n = \frac{c}{n^k}$ for some positive integer $k$ and a positive constant $c$. We will explore what value of $k$ would make the sequence satisfy both properties.
For property a): $ na_n = n \cdot \frac{c}{n^k} = c \cdot n^{1-k} $ For $c \cdot n^{1-k}$ to be a positive integer, $1 - k$ must be zero, meaning $k = 1$.
So, $a_n = \frac{c}{n}$ seems to be the correct form for satisfying property (a).
Now, let's check if this satisfies property (b). $$ \sum_{k=1}^{n} (-1)^{k-1} a_k \binom{n}{k} = \sum_{k=1}^{n} a_k. $$ Substituting $a_k = \frac{c}{k}$: $$ \sum_{k=1}^{n} (-1)^{k-1} \left(\frac{c}{k}\right) \binom{n}{k} = \sum_{k=1}^{n} \frac{c}{k}. $$ This is the same equation we started with. Since we already proved that $a_n = \frac{c}{n}$ satisfies property a), we conclude that it also satisfies property b).
Therefore, the general term of the sequence $(a_n)_{n\ge1}$ is $a_n = \frac{c}{n}$, where $c$ is a positive constant.
Source: Romanian National Olympiad 2022 shortlist problem