Finding f(n) for All Multiples of 5

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Dustinsfl
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F is all integers that are multiplies of 5

[tex]f(n) = <br /> \begin{cases} <br /> 0, & \mbox{if }n\mbox{ 1} \\<br /> \frac{5n}{2}, & \mbox{if }n\mbox{ is even}\\<br /> ? , & \mbox{if }n\mbox{ is odd}<br /> \end{cases}[/tex]

N exists in the natural number
I don't know how to obtain the function that will represent the odd integers.
 
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Are you trying to get a function that produces all the integer multiples of 5?

Whats wrong with f(n) = 5n ?
 
Gib Z said:
Are you trying to get a function that produces all the integer multiples of 5?

Whats wrong with f(n) = 5n ?

I forgot to mention n exists in the natural numbers.
 
Ok then, the n is even part of your function will give all the positive multiples, now try something similar for n odd. I don't know how to give you a decent hint without giving it away, but basically, the odd case is very similar to the even case, except with a negative sign, and a little change to the n term to make sure it's divisible by 2.