11.4. The sequence of natural numbers $a_{n}, n=1,2, \ldots$ is such that $a_{1}=1$ and $a_{n}=\frac{n+1}{n-1}\left(a_{1}+a_{2}+\ldots+a_{n-1}\right)$ for all $n=2,3, \ldots$. Find the formula for the "general term" of the sequence, that is, a formula that explicitly expresses $a_{n}$ in terms of $n$ for any $n$.
Answer. $a_{n}=(n+1) 2^{n-2}$

Solution. Let's find the first few terms of the sequence: $a_{1}=1, a_{2}=\frac{3}{1} \cdot 1=3 \cdot 1$, $a_{3}=\frac{4}{2} \cdot(1+3)=8=4 \cdot 2, a_{4}=\frac{5}{3} \cdot(1+3+8)=20=5 \cdot 4$ and make the assumption that $a_{n}=(n+1) 2^{n-2}$. We also calculate the sums of the first terms of this sequence: $s_{1}=a_{1}=1, \quad s_{2}=a_{1}+a_{2}=1+3=4=2 \cdot 2, \quad s_{3}=a_{1}+a_{2}+a_{3}=1+3+8=12=3 \cdot 4$ $s_{4}=a_{1}+a_{2}+a_{3}+a_{4}=1+3+8+20=32=4 \cdot 8$, and make the assumption that $s_{n}=n \cdot 2^{n-1} \cdot$

Let's prove the made assumptions by mathematical induction. First, $a_{n+1}=\frac{n+2}{n} \cdot s_{n}=\frac{n+2}{n} \cdot n \cdot 2^{n-1}=(n+2) \cdot 2^{n-1}=((n+1)+1) \cdot 2^{(n+1)-2}$ is true. Next, $s_{n+1}=s_{n}+a_{n+1}=n \cdot 2^{n-1}+(n+2) \cdot 2^{n-1}=2(n+1) \cdot 2^{n-1}=(n+1) \cdot 2^{(n+1)-1}$ - also true.