4, 1. To join a club, each new member must pay each member of the club the same amount that he had to pay in total when he joined plus one euro. If the first member paid one euro, how much will the $n$-th member have to pay in total?
Solution. Let $a_{n}$ be the total quota of the $n$-th member and let $s_{n}=a_{1}+\cdots+a_{n}$. The $n$-th $(n \geq 2)$ member has to pay in total $\left(a_{1}+1\right)+\left(a_{2}+1\right)+\cdots+$ $\left(a_{n-1}+1\right)=s_{n-1}+n-1$ euros, so $a_{n}=s_{n-1}+n-1$ and

$$
s_{n}=s_{n-1}+a_{n}=s_{n-1}+s_{n-1}+(n-1)=2 s_{n-1}+n-1
$$

Iterating this relation yields $s_{n}=2^{n-1}+2^{n-2} \times 1+2^{n-3} \times 2+\cdots+2 \times(n-$ 2) $+(n-1)$, from which $s_{n}=2 s_{n}-s_{n}=2^{n}+2^{n-2}+2^{n-3}+\cdots+2-n+1=$ $2^{n}+2^{n-1}-1-n$ and then, for $n \geq 2$,

$$
a_{n}=s_{n}-s_{n-1}=2^{n}-2^{n-2}-1=3 \times 2^{n-2}-1
$$