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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.

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2 Answers 2

up vote 3 down vote accepted

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

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Wow, it's so simple when one can see it. Is this just something you know? –  Neil Barnwell Jul 25 '11 at 15:49
    
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. –  The Chaz 2.0 Jul 25 '11 at 15:52
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 $-1$ –  Ross Millikan Jul 25 '11 at 15:53
1  
Ross Thanks so much for your swift answer. @The Chaz - thank you for your helpful explanation. –  Neil Barnwell Jul 25 '11 at 15:57
    
@Neil: Anytime! –  The Chaz 2.0 Jul 25 '11 at 16:04

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

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