summation - Sum of first $n$ triangle numbers, without induction

Background

I wish to calculate $$ S= \sum_{i = 1}^{n}\frac{k(k+1)}{2}$$

I know what the answer is going to be, since this is essentially the sum of the first $n$ triangle numbers.

I.e. $S = (1) + (1+2) + (1+2+3) + \cdots + (1+2+3+\cdots+n)$

All solutions I've seen seem to know in advance what the answer is going to be, and their problem is proving it, which can be done using induction.

Question

However, I was wondering if this can be calculated using formulae for summations instead?

Alternatively, for instances where we do wish to prove it instead of calculate it, are there any other ways besides induction?

For reference

The answer should be $$\frac{n(n+1)(n+2)}{6}$$

Answer

If you know:

$$
\sum\limits_{k=1}^n k = \frac{n(n+1)}{2}
$$
and
$$
\sum\limits_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
$$
then:
$$
\sum\limits_{k=1}^n \frac{k(k+1)}{2} = \frac{1}{2} \left[\sum\limits_{k=1}^n k^2 + \sum\limits_{k=1}^n k \right] = \frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right] = \frac{n(n+1)(n+2)}{6}

$$