 # Pattern: 1 2 6 14 27 46 72

• Mar 26th, 2020, 05:01 AM
dee-u
Pattern: 1 2 6 14 27 46 72
What could be the formula for the following sequence of numbers: 1 2 6 14 27 46 72
• Mar 26th, 2020, 07:48 AM
passel
Re: Pattern: 1 2 6 14 27 46 72
What do you think the formula should look like?
My initial stab at generating the series with code would be (using VB6):
Code:

```Option Explicit Private Sub Command1_Click()   Dim a As Integer, b As Integer, c As Integer, d As Integer   a = 1   b = 2   Debug.Print a   Debug.Print b   For c = 3 To 7     d = (b - a) + c     a = b     b = b + d     Debug.Print b   Next End Sub```
I don't know if that helps with determining a formula, or whether the above code could be simplified.
• Mar 26th, 2020, 09:15 AM
OptionBase1
Re: Pattern: 1 2 6 14 27 46 72
First step is to see if the differences of terms gives any insight into any formula used to generate the terms:

1 2 6 14 27 46 72

1st differences: 1 4 8 13 19 26
2nd differences: 3 4 5 6 7
3rd differences: 1 1 1 1

Since the 3rd differences are all equal, that means the formula used to generate the terms is of 3rd degree in the form of:

A(x) = a*x^3 + b*x^2 + c*x + d

You have been given A(1) = 1, A(2) = 2, A(3) = 6, A(4) = 14, ...

Using those given values, you can build a series of simultaneous equations that allow you to solve for the values of the coefficients a, b, c, and d.

a + b + c + d = 1
8*a + 4*b + 2*c + d = 2
27*a + 9*b + 3*c + d = 6
64*a + 16*b + 4*c + d = 14

That's as far as I'll go for now. Good luck!