Thread: Modelling tetrahedral numbers

1. Modelling tetrahedral numbers

Hi,

I hope someone can help. I'm trying to solve for a):

I developed the equation y = 4^x, where "x" represents an additional 4 sides. This equation works perfectly well, however, it is different from the textbook's solution. The textbook's solution is the following:

Can someone please explain the reasoning behind this equation? It seems pretty complex, and it would be great if someone could provide me with a step-by-step process of the thinking of this equation.

2. Re: Modelling tetrahedral numbers

Huh? I do not understand your idea at all. What does $y=4^x$ have to do with the number of dots?

The step-by-step process is to find differences between numbers. You create a reverse number pyramid to find differences between numbers. You keep going until you reach all zeros. The number of rows minus two is the degree of the polynomial needed to represent the equation.

$\begin{matrix}1 & & 4 & & 10 & & 20 & & 35 \\ & 3 & & 6 & & 10 & & 15 & \\ & & 3 & & 4 & & 5 & & \\ & & & 1 & & 1 & & & \\ & & & & 0 & & & & \end{matrix}$

There are five rows, so you need a third degree polynomial.

$T(n) = an^3+bn^2+cn+d$

From the diagram:
$T(1) = 1$
$T(2) = 4$
$T(3) = 10$
$T(4) = 20$

Plugging in:

$1 = a+b+c+d$ (Equation 1)
$4 = 8a+4b+2c+d$ (Equation 2)
$10 = 27a+9b+3c+d$ (Equation 3)
$20 = 64a+16b+4c+d$ (Equation 4)

Now, we solve the system of four equations.
Subtract the first equation from the next three:
$3 = 7a+3b+c$ (Equation 5)
$9 = 26a+8b+2c$ (Equation 6)
$19 = 63a+15b+3c$ (Equation 7)

Subtract 2x Equation 5 from Equation 6 and 3x Equation 5 from Equation 7:
$3 = 12a+2b$ (Equation 8)
$10 = 42a+6b$ (Equation 9)

Subtract 3x Equation 8 from Equation 9:
$1 = 6a$

This gives $a = \dfrac{1}{6}$. We plug that into either Equation 8 or Equation 9. Let's plug it into Equation 8:
$3 = 2+2b \Longrightarrow b = \dfrac{1}{2}$

Now, we plug both $a$ and $b$ into Equation 5:
$3 = \dfrac{7}{6}+\dfrac{3}{2}+c \Longrightarrow c = \dfrac{2}{6} = \dfrac{1}{3}$

Finally, we plug all four into equation #1 to get $d$:
$1 = \dfrac{1}{6}+\dfrac{1}{2}+\dfrac{1}{3}+d \Longrightarrow d=0$

Thus, $T(n) = \dfrac{1}{6}n^3+\dfrac{1}{2}n^2+\dfrac{1}{3}n$

There are many ways of finding this pattern. This is just one of many.

3. Re: Modelling tetrahedral numbers

If by "x" you mean what I would call "layers", $4^x$ clearly does NOT work. $4^0= 1$ and $4^1= 4$ but $4^2= 16$ not 10, $4^3= 64$, not 20, and $4^4= 256$, not 35.

4. Re: Modelling tetrahedral numbers

Originally Posted by SlipEternal
Huh? I do not understand your idea at all. What does $y=4^x$ have to do with the number of dots?

The step-by-step process is to find differences between numbers. You create a reverse number pyramid to find differences between numbers. You keep going until you reach all zeros. The number of rows minus two is the degree of the polynomial needed to represent the equation.

$\begin{matrix}1 & & 4 & & 10 & & 20 & & 35 \\ & 3 & & 6 & & 10 & & 15 & \\ & & 3 & & 4 & & 5 & & \\ & & & 1 & & 1 & & & \\ & & & & 0 & & & & \end{matrix}$

There are five rows, so you need a third degree polynomial.

$T(n) = an^3+bn^2+cn+d$

From the diagram:
$T(1) = 1$
$T(2) = 4$
$T(3) = 10$
$T(4) = 20$

Plugging in:

$1 = a+b+c+d$ (Equation 1)
$4 = 8a+4b+2c+d$ (Equation 2)
$10 = 27a+9b+3c+d$ (Equation 3)
$20 = 64a+16b+4c+d$ (Equation 4)

Now, we solve the system of four equations.
Subtract the first equation from the next three:
$3 = 7a+3b+c$ (Equation 5)
$9 = 26a+8b+2c$ (Equation 6)
$19 = 63a+15b+3c$ (Equation 7)

Subtract 2x Equation 5 from Equation 6 and 3x Equation 5 from Equation 7:
$3 = 12a+2b$ (Equation 8)
$10 = 42a+6b$ (Equation 9)

Subtract 3x Equation 8 from Equation 9:
$1 = 6a$

This gives $a = \dfrac{1}{6}$. We plug that into either Equation 8 or Equation 9. Let's plug it into Equation 8:
$3 = 2+2b \Longrightarrow b = \dfrac{1}{2}$

Now, we plug both $a$ and $b$ into Equation 5:
$3 = \dfrac{7}{6}+\dfrac{3}{2}+c \Longrightarrow c = \dfrac{2}{6} = \dfrac{1}{3}$

Finally, we plug all four into equation #1 to get $d$:
$1 = \dfrac{1}{6}+\dfrac{1}{2}+\dfrac{1}{3}+d \Longrightarrow d=0$

Thus, $T(n) = \dfrac{1}{6}n^3+\dfrac{1}{2}n^2+\dfrac{1}{3}n$

There are many ways of finding this pattern. This is just one of many.

This was a great explanation. Thank you

5. Re: Modelling tetrahedral numbers

Lol good point. SlipEternal's explanation helped.