0
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Which "classic proof" of $\sum_{j=1}^nj^2=\frac{n(n+1)(2n+1)}6$? $\endgroup$ – Simply Beautiful Art Dec 3 '16 at 18:54
  • $\begingroup$ 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. $\endgroup$ – dmitzov Dec 3 '16 at 19:00
  • $\begingroup$ Personally, I think that method is very tedious in comparison to my below answer. But do what you will. $\endgroup$ – Simply Beautiful Art Dec 3 '16 at 19:03
0
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.