relatively simple sequence

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relatively simple sequence

Postby Dmitry » Wed May 16, 2012 2:00 pm UTC

Can somebody please explain how to derive that the explicit formula for
-1, 5, -9, 13, -17, ...

is a sub n = [(-1)^n](4n-3), n >= 1; or a sub n = [(-1)^(n+1)](4n+1)

I do not understand how it can be done :|

Thanks ahead of time!
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Re: relatively simple sequence

Postby gmalivuk » Wed May 16, 2012 2:30 pm UTC

The only difference there is whether you consider the first element (-1) to be a0 or a1.
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Re: relatively simple sequence

Postby Dmitry » Wed May 16, 2012 3:34 pm UTC

I understand the part of (-1)^n and (-1)^(n+1)

what i do not understand is, what follows that.

why 4n-3, and why 4n+1
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Re: relatively simple sequence

Postby Sizik » Wed May 16, 2012 4:47 pm UTC

Taking the absolute value of everything, the series is 1, 5, 9, 13, 17, 21, etc.. If you take the differences between successive elements, you get 4, 4, 4, 4, etc., which tells you that the series can be expressed as 4*0 + 1, 4*1 + 1, 4*2 + 1, etc. An easy way to express this would be an = 4n + 1, having the first element, 1, as a0. If you wanted 1 to be a1 instead, then you would have an = 4(n-1) + 1 = 4n + 1 - 4 = 4n - 3. Afterwards, just add in the (-1)n or (-1)n+1, depending on which way you chose to describe the series.
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Re: relatively simple sequence

Postby Dmitry » Wed May 16, 2012 5:13 pm UTC

Alright that cleared it up!

Thanks =]
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