The sum of the first n tetrahedral numbers

[showmyiq]

What is tetrahedral numbers?

Well you can see this puzzle published on 1st September 2012:
http://www.showmyiq.com/2012/September/ ... iddle.html

Tetrahedral number in fact counts the balls in one tetrahedral pyramid.

The first few tetrahedral numbers are:
1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455, 560, 680, 816, 969

But how to found the sum of the first n?

Well first we need to found an equation describing each pyramid.
We can observe that one pyramid is based on layers.

Let’s define the first layer, starting from the top as layer 1.
We can see that in each pyramid layer 1 is always the same – 1 ball.
We get the second layer by counting the balls of the first layer and adding 2 balls.
We get the third layer by counting the balls of the second layer and adding 3 balls.
See the pattern?
We get the nth layer by counting the balls of the previous (n-1)th layer and adding n balls.

So if we need to describe this equation in math words we get this:

Layer(n) = n + Layer(n-1)

What about Pyramid(n)?
Well Pyramid(n) = Layer(n) + Layer(n-1) + … + Layer(2) + Layer(1)

How many balls one Layer has? Well as you can see each Layer has rows and rows are arranged by arithmetic progression of 1. I mean the count of the balls in each layer is the sum of the first n natural numbers, if the layer is on the nth position.Example:

Layer(1) has 1 balls
Layer(2) has 1+2 = 3 balls
Layer(3) has 1+2+3 = 6 balls
….
Layer(n) has 1+2+3 + … + n = n*(n+1)/2 balls

You can see the proof of that formula here:
The Sum of the first n natural numbers

So in fact we just discovered something pretty interesting.

Pyramid(n) = Layer(n) + Layer(n-1) + … + Layer(2) + Layer(1)
\begin{equation}
Pyramid_n =
(1+2+3 + .. +n ) +
(1+2+3 + …(n-1)) +
… +
(1+2+3)+
(1+2) +
1
\end{equation}

We can see that we have exactly n times 1.
Exactly (n-1) times 2. Exactly (n-2) times 3 and etc.

So all this sum is equal in fact to this:
Pyramid(n) = n.1 + (n-1).2 + .. + 2.(n-1) + n

But if we don’t make this calculation, but just writing down the previous equation in more math-friendly language, we get this:
\begin{equation}
Pyramid_n = \sum_{i=1}^{n} \frac {i(i+1)} {2} \\
Pyramid_n = \sum_{i=1}^{n} \frac {i^2} {2} +\sum_{i=1}^n\frac {i} {2}\\
\end{equation}

Both of the sums in the right side of the equation we already solved in previous topics. This in facts are the sum of all the squares of the first n natural numbers and the sum of all the first n natural numbers.
For the sums of all the squares of the first n natural numbers proof you can follow this link:
The sum of the first n squares of natural numbers

So far – so good!

Now let’s see what is going on!

\begin{equation}
Pyramid_n = \frac {1} {2}( \sum_{i=1}^{n} \frac {i^2} {2} +\sum_{i=1}^n\frac {i} {2})\\
Pyramid_n = \frac {1} {2}\sum_{i=1}^{n}i^2 + \frac{1}{4}\sum_{i=1}^{n}i\\
Pyramid_n =\frac {n(n+1)(2n+1)} {12} + \frac {n(n+1)} {4}\\
Pyramid_n =\frac {n(n+1)(2n+1)} {12} + \frac {3n(n+1)} {12}\\
Pyramid_n =\frac {n(n+1)(2n+4)} {12}\\
Pyramid_n =\frac {n(n+1)( n+2)} {6}\\
\end{equation}

Eureka!
Let the apples hit us!

35 balls in tetrahedral pyramid

35 balls Pyramid.gif (188.97 KiB) Viewed 8573 times