1. Determine the functions $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$, if $f(1)-2 f(2)+3 f(3)-4 f(4)+\ldots+(-1)^{n-1} n f(n)=\frac{(-1)^{n-1} n f(n+1)}{2}$, for any natural number $n \in \mathbb{N}^{*}$.

Dan Popescu, Suceava
Solution: If $f(1)=a \in \mathbb{N}^{*}$, for $n=1$ in the statement, it follows that $f(2)=2a$.

Since $f(1)-2f(2)+3f(3)-4f(4)+\ldots+(-1)^{n-2}(n-1)f(n-1)=\frac{(-1)^{n-2}(n-1)f(n)}{2}, \forall n \in \mathbb{N}^{*}$, with the relation from the statement,

it follows that $\frac{(-1)^{n-1} n f(n+1)}{2}=(-1)^{n-1} n f(n)+\frac{(-1)^{n-2}(n-1) f(n)}{2}, \forall n \in \mathbb{N}^{*}$. We find $(n+1) f(n)=n f(n+1), \forall n \in \mathbb{N}^{*}$.

Thus, under the inductive hypothesis $f(n)=a n$, the last relation ensures $f(n+1)=a(n+1)$, which leads to $f(n)=a n$,

$\forall n \in \mathbb{N}^{*}$.

Scoring:

| Deduce $f(2)$ in terms of $f(1)$ | $1 \mathrm{p}$ |
| :--- | :--- |
| Find the relationship between $f(n)$ and $f(n+1)$ | $2 \mathrm{p}$ |
| Prove inductively that $f(n)=n f(1), \forall n \in \mathbb{N}$ | $4 \mathrm{p}$ |