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.