0
$\begingroup$

I have the following sequences that form a relationship:

x: 1 2 3 4 5
y: 1 2 4 8 16

Where each number in the second sequence is twice the previous number.

Is there a formula that will give me the correct value of y for a given value of x? Does it have a name, and how could I have gone about working this out for myself?

Many thanks in advance.

2 Answers 2

3

You have $y=2^{x-1}$ and some characters

  • 0
    Wow, it's so simple when one can see it. Is this just something you *know*?2011-07-25
  • 0
    When the ratio of consecutive terms is constant (2, in this case), it is called geometric, and has the form $y = x_0r^{x-h}$, where $x_0$ is the first term and $r$ is the ratio.2011-07-25
  • 2
    @Neil: When you say each number is double the previous, that makes $2^n$ come to mind. But checking that is off by a factor $2$, which is corrected by the $-1$2011-07-25
  • 1
    Ross Thanks so much for your swift answer. @The Chaz - thank you for your helpful explanation.2011-07-25
  • 0
    @Neil: Anytime!2011-07-25
  • 0
    @Neil: The expression from The Chaz has two constants, $x_0$ and $h$, which are redundant. Two versions where $x_0'r^{-h'}=x_0r^{-h}$ will give the same sequence.2011-07-25
  • 0
    @Ross: It is possible to work around the character limit by hiding the padding in a HTML comment, like ``.2011-07-26
1

$y = 2^{x-1}$. This is a geometric sequence.