How to prove that $\sum\limits_{j = 1}^{n} j(j + 1) = \frac{1}{3}n(n + 1)(n + 2)\ \text{where }\ n \geq 1.$ I would like to know how it is to prove this problem:
$$\sum_{j = 1}^{n} j(j + 1) = \frac{1}{3}n(n + 1)(n + 2)  \text{where } n \geq 1 \\
$$

Observe that $$\sum\limits_{j=1}^{n} j^2 = \frac {1} {6}  n(n+1) (2n+1)$$ and $$\sum\limits_{j=1}^{n} j = \frac {n(n+1)} {2}.$$
EDIT $:$ 
To prove the first sum consider the equality $(t+1)^3 - t^3 = 3t^2 + 3t +1.$ Take $t=1,2, \cdots, n.$ Then we get by summing the equality for $t=1,2, \cdots,n$ $$3\sum\limits_{t=1}^{n} t^2 = \sum\limits_{t=1}^{n} ((t+1)^3 - t^3) - 3 \left ( \sum\limits_{t=1}^{n} t \right ) - n.$$
Can you proceed now?
What is $\sum\limits_{t=1}^{n} ((t+1)^3 - t^3)$?
Observe that  $\sum\limits_{t=1}^{n} ((t+1)^3 - t^3) = (n+1)^3 -1.$ Also you know that $\sum\limits_{t=1}^{n} t = \frac {n(n+1)} {2}.$ Now you should plug all these things to get the required answer.
So $$\sum\limits_{t=1}^{n}t^2 = \frac 1 3 \left ((n+1)^3 - 1 - \frac {3n(n+1)} {2}  - n \right ).$$
I leave the further simplifications to you.
