186. The sequence $\left(a_{n}\right)$ is defined by the conditions:

$$
a_{1}=0, \quad a_{n+1}=a_{n}+n
$$

Find the formula for the general term of the sequence.
$\triangle$ Let's represent the second condition in the following form:

$$
a_{k+1}=a_{k}+k, \quad a_{k+1}-a_{k}=k
$$

Let's set $k=1,2,3, \ldots, n-1$ in the last equality. We will obtain a series of equalities, which we will then add term by term:

$$
\begin{aligned}
& a_{2}-a_{1}=1 \\
& a_{3}-a_{2}=2, \\
& +a_{4}-a_{3}=3 \\
& a_{n}-a_{n-1}=n-1 \\
& a_{n}=1+2+3+\ldots+(n-1)=\frac{n(n-1)}{2}
\end{aligned}
$$

(Here, the formula for the sum of the first $n$ terms of an arithmetic progression was applied.)

Answer: $a_{n}=\frac{n(n-1)}{2}$.