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: 
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.
Because $n-1=\frac{a\_{n}-a\_{n-1}}{n-(n-1)} \approx \frac{d}{dn} a\_n$ is linear, it is reasonable to guess that $a\_n$ is a quadratic because the integral of a linear function is a quadratic.

So guess $a\_n=an^2+bn+c$.

This means,

$$\left(an^2+bn+c \right)-\left(a(n-1)^2+b(n-1)+c \right)$$

$$=a(2n-1)+b$$

$$=\color{red}{2a}n+\color{blue}{b-a}$$

$$:=\color{red}{1}n\color{blue}{-1}$$

We want:

$$2a=1$$

$$b-a=-1$$

$$a(1^2)+b(1)+c=0$$