Question: Given sequence:

$A\_1 = 0$
$A\_2 = 1$
$A\_3 = 3$
$A\_4 = 6$
$A\_5 = 10$ ...

And the recursive formula is given as $A\_n = A\_{n-1} + (n - 1)$

How can one find this sequence's explicit formula?

Answer: 
**Hint**. One may write
$$
A\_n - A\_{n-1} = (n - 1),\qquad n\ge2,
$$ then summing from $2$ to $N$ terms *telescope on the left hand side* getting
$$
A\_N - A\_1 = \sum\_{n=2}^N(n - 1)=\frac{N(N-1)}2.
$$
We recognise $0,1,3,6,10,\cdots$ as $\displaystyle \left[\binom 12\right], \binom 22, \binom 32,\binom42, \binom 52,\cdots $
and conjecture that $$\color{red}{A\_n=\binom n2}$$
Testing this in the recurrence relation given:
$$A\_n-A\_{n-1}=\binom n2-\binom {n-1}2=\frac {n(n-1)}2-\frac {(n-1)(n-2)}2=n-1\\
\Longrightarrow A\_n=A\_{n-1}+(n-1)$$
Hence conjecture is true.