How did we derive this general term for the series? We have this series of numbers:
$1, 3, 6, 10, 15$
The general term can be described wit: $\frac{r(r + 1)}{2}$
Apparently the following series:
$1, 4, 10, 20, 35$
Can be described with $\frac{r(r + 1)(r + 2)}{6}$ based on the first series.
But I am not clear how this is derived. Can someone please explain?

$S = 1 + 4 +10 + 20 + 35 \cdots + a_{n-1} + a_n  \cdots - 1$
$S = 1 + 4 +10 + 20 + 35 \cdots + a_{n-1} + a_n \cdots - 2$
Subtracting 1 from 2
$0 = 1 + ((4 - 1) + (10 - 4) + (20 - 10) + (35 - 20)+ \cdots +  a_{n} - a_{n-1}) - a_n$
$a_n = 1 + ((4 - 1) + (10 - 4) + (20 - 10) + (35 - 20)+ \cdots +  a_{n} - a_{n-1})$
$a_n = 1 + 3 + 6 + 10 + 15 + \cdots +  (a_{n} - a_{n-1})$
$a_n = \sum_{r = 0}^n \frac{r(r + 1)}{2}$
$a_n = \frac{1}{2} * \sum_{r = 0}^n (r^2 + r)$
$a_n = \frac{1}{2} * (\sum_{r = 0}^n r^2 + \sum_{r = 0}^nr)$
$a_n = \frac{1}{2} * (\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2})$
$a_n = \frac{n(n+1)}{2} * (\frac{(2n+1)}{6} + \frac{1}{2})$
$a_n = \frac{n(n+1)}{2} * (\frac{(2n+1 +3)}{6})$
$a_n = \frac{n(n+1)}{2} * (\frac{(2n+4)}{6})$
$a_n = \frac{n(n+1)}{2} * (\frac{(n+2)}{3})$
$a_n = \frac{n(n+1)(n+2)}{6}$
This way is pretty easy.
