4. Find all natural numbers $n \geq 2$, for which the equality $4 x_{n}+2 y_{n}=55 n^{2}+61 n-116$ holds, where $x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n, y_{n}=1^{2}+2^{2}+3^{2}+\cdots+(n-1)^{2}$.

(20 points)
Solution: Let $z_{n}=1+2+\cdots+n=\frac{(n+1) n}{2}=\frac{n^{2}+n}{2}$. We have

$$
\begin{aligned}
& x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n=(2+3+\cdots n)+(3+4+\cdots n)+\cdots((n-1)+n)+n= \\
& =\left(z_{n}-z_{1}\right)+\left(z_{n}-z_{2}\right)+\cdots+\left(z_{n}-z_{n-1}\right)=(n-1) z_{n}-\left(z_{1}+z_{2}+\cdots+z_{n-1}\right)= \\
& =\frac{1}{2}\left((n-1) n(n+1)-\left(1^{2}+1+2^{2}+2+\cdots+(n-1)^{2}+(n-1)\right)=\right. \\
& =\frac{1}{2}\left((n-1) n(n+1)-\left(y_{n}+z_{n-1}\right)\right)
\end{aligned}
$$

Substitute the obtained expression into the equation $4 x_{n}+2 y_{n}=55 n^{2}+61 n-116$. We get

$$
\begin{aligned}
& 2\left((n-1) n(n+1)-\left(y_{n}+z_{n-1}\right)\right)+2 y_{n}=55 n^{2}+61 n-116,2(n-1) n(n+1)-n(n-1)=55 n^{2}+61 n-116 \\
& (n-1)(2 n(n+1)-n)=(n-1)(55 n+116), 2 n^{2}-54 n-116=0, n^{2}-27 n-58=0, n=29
\end{aligned}
$$

Answer: $n=29$.