Finding the partial sum of n(n+1) How do I prove that $\sum_{j=1}^n j(j+1) = \frac 13n(n+1)(n+2)$ without using induction?
Is there a proof like the classic one for $\sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$?
 A: Use physics:
The body consists of equal particles arranged into a triangle:

(Particles are placed in a squared net, so the horizontal distance between adjacent columns is $1$.)
Now compute its torque (moment of force = distance $\times$ force) caused by the gravitational force, related to the leftmost (red) point by two methods:
Method 1. As the sum of torques of its parts: Add torques of individual particles (grouped by columns).

The result is $$\ 0 \cdot 1\ +\ 1\ \cdot 2\ + \ 2 \cdot 3\ + \ \dots + \ n(n+1).$$
 
Method 2. As torque of the whole body: Multiply the distance to the center of its mass by the gravitational force of the whole body:

Since the gravitational force of the whole body is the sum of gravitational forces of all its particles, the  result is $$\left(\frac {2n} 3\right) \cdot (1 + 2 + \dots + n + 1) = \frac 2 3 n \frac {(n+1)(n+2)} 2 = \frac 1 3 n(n+1)(n+2).$$
As both method must give the same result, you obtain the given formula.
A: Consider $$f(r)=r(r+1)(r+2)$$
After simplification, $$f(r)-f(r-1)=3r(r+1)$$
Now apply a $\Sigma$ sign to both sides and we have a telescoping series which gives $$n(n+1)(n+2)=3\sum_{r=1}^n r(r+1)$$ and the result follows.
A: Recall the Hockey-stick identity.  It can easily be seen from it that
$$\sum_{j=1}^n\frac{j(j+1)}2=\frac{n(n+1)(n+2)}6$$
Multiplying both sides by $2$, we get
$$\sum_{j=1}^nj(j+1)=\frac{n(n+1)(n+2)}3$$
In general, the Hockey-stick identity lets us say that
$$\sum_{j=1}^nj(j+1)(j+2)\dots(j+p)=\frac{n(n+1)(n+2)\dots(n+p)(n+p+1)}{p+2}$$
