50. The sequence of numbers $1,4,10,19, \ldots$ has the property that the differences between two consecutive numbers form an arithmetic progression. Find the $n$-th term and the sum of the first $n$ terms of this sequence.
50. Let the terms of the arithmetic progression be $b_{1}, b_{2}, b_{3}, \ldots, b_{n}$, and the terms of the sequence be $a_{1}, a_{2}, a_{3}, \ldots$. According to the condition, $b_{1}=3, b_{2}=6$. Therefore, $d=b=3$. Then

$a_{2}-a_{1}=b$

$a_{3}-a_{2}=2 b$

$a_{4}-a_{3}=3 b$

$\dot{a}_{n}-\dot{a}_{n-1}=(n-1) b$

By adding term by term, we get the general term of the sequence $a_{n}=1+\frac{3}{2} n(n-1)$.

Then $S_{n}=a_{1}+a_{2}+a_{3}+\ldots+a_{n}=n+\frac{3}{2}[1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+(n-1) n]$.

But $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+(n-1) n=\frac{(n-1) n(n+1)}{3}$. Therefore, $S_{n}=\frac{1}{2} n\left(n^{2}+1\right)$.