# 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 $\sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$?

• Which "classic proof" of $\sum_{j=1}^nj^2=\frac{n(n+1)(2n+1)}6$? – Simply Beautiful Art Dec 3 '16 at 18:54
• I'll try to make it in edit to this reply: new to this place and latex and a novice. It's the one where you start by taking $(k+1)^3 = k^3 + 3k^2 + 3k +1$, subtract k^3 from both sides, take the sum of it for n = (1,2,3,...,n) and do some algebra. – dmitzov Dec 3 '16 at 19:00
• Personally, I think that method is very tedious in comparison to my below answer. But do what you will. – Simply Beautiful Art Dec 3 '16 at 19:03

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

$$\sum_{j=1}^nj(j+1)(j+2)\dots(j+p)=\frac{n(n+1)(n+2)\dots(n+p)(n+p+1)}{p+2}$$

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.

Use physics: The body is comprised of equal particles arranged into a triangle:

              o
o
o     o
o     o
o     o     o
o     o
o     o
o
o


(Particles are placed in 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 point by these two methods:

1. As the sum of torques of its parts: Add torques of individual particles (grouped by columns).

2. For the whole body: Multiply the distance to the center of its mass by the gravitational force.

Method 1:     $$0 \cdot 1 + 1 \cdot 2 + \dots + n(n+1)$$

Method 2:   $$(\frac 2 3 n) \cdot (1 + 2 + \dots + n + 1) = \frac 2 3 n \frac {(n+1)(n+2)} 2 = \frac 1 3 n(n+1)(n+2)$$

As the resul must be the same, you obtain the given formula.