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## Finite Sum - Modified Geometric Series

Does anyone know how to evaluate

$$S_n = \sum_{i=0}^{n-1} i2^i$$

I tried the following. Let r = 2, and figure out the terms in

$$S_n - rS_n$$

Unlike with a regular geometric series, this does not make all but two of the terms disappear. But it does make all but one of the terms turn into a simple power of 2 (once you collect like powers of 2). In other words, it turns into something plus a regular geometric series. For my final answer, solving for S_n, I got:

$$S_n = (n-2)2^n + 2$$

but I have reason to believe this is incorrect. Can anybody help me out?
 Recognitions: Homework Help Science Advisor I think you're right. Check yourself against A036799 if you like (but watch the offset!).
 Recognitions: Science Advisor Using elemtary calculus I got (n-1)2n+2

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## Finite Sum - Modified Geometric Series

Perhaps, it is unnecessary but one way is to use the properties of the z transform. The sum could be equal to the z transform with z=1. I’ll check later. There are two time domain operations performed on the infinite geometric series. They are differentiation and multiplication by a rec function. These operations have equivalents in the z domain.

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 Quote by CRGreathouse I think you're right. Check yourself against A036799 if you like (but watch the offset!).
The online encyclopedia of integer sequences? I had no idea that it existed. My answer looks consistent with theirs.

Let n = 4, and use their formula (we should get 98)

$$2+2^5(4-1) = 2+32(3) = 2+ 96 = 98$$

Let n-1 = 4 and use MY formula (we should get 98)

$$2 + 2^n(n-2) = 2 + 2^5(3) = 98$$

Edit: That confirms my answer. Thanks for the link CRGreathouse!

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 Quote by mathman Using elemtary calculus I got (n-1)2n+2
Exactly what method did you use?

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 Quote by John Creighto Perhaps, it is unnecessary but one way is to use the properties of the z transform. The sum could be equal to the z transform with z=1. I’ll check later. There are two time domain operations performed on the infinite geometric series. They are differentiation and multiplication by a rec function. These operations have equivalents in the z domain.
Right, from what I can remember, the z-transform of a sequence a(n) is a function of a complex variable f(z) such that {a(n)} are just the coefficients (in reverse order) of the Laurent series expansion of f(z) (about zero?)

$$f(z) = \sum_{n=-\infty}^{\infty} a(n)z^{-n}$$

And you were thinking of the property that:

$$\frac{d}{dz}f(z) = - \sum_{n=-\infty}^{\infty} na(n)z^{-n-1} = - z^{-1}\sum_{n=-\infty}^{\infty} na(n)z^{-n} = -z^{-1}\mathcal{Z}\{na(n)\}$$

Or in other words

$$\mathcal{Z}\{na(n)\} = -z\frac{d}{dz}\mathcal{Z}\{a(n)\}$$

And so my series would be:

$$-(z)\frac{d}{dz}\mathcal{Z}\{1\}$$

evaluated at

$$z = 2^{-1}$$

EDIT: But I just realized that these are INFINITE series, and I'm dealing with a FINITE series, so I don't think that any of this is relevant. AARRGH!

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 Quote by cepheid But I just realized that these are INFINITE series, and I'm dealing with a FINITE series, so I don't think that any of this is relevant. AARRGH!
Any finite series can be extended into an equal infinite series....

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 Quote by Hurkyl Any finite series can be extended into an equal infinite series....
Right, of course. I was being stupid. Of course you can take the z transform of a finite sequence. I've done it many times. Formally, you let a(i) = 1 for 0 <= i <= n-1, and a(i) = 0 everywhere else.

Then the sum you evaluate gives you:

$$\mathcal{Z}\{a(i)\} = \frac{1-z^{-n}}{1-z^{-1}}$$

Then you differentiate this wrt z and then you multiply that by -z and then you plug in z = 1/2. I did that and got the same answer as in my original post.

So the answer has been verified by two separate methods.

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 Quote by cepheid Exactly what method did you use?
f(x)=sum(0,n-1)xk=(xn-1)/(x-1)

f'(x)=sum(0,n-1)kxk-1={(x-1)nxn-1-(xn-1)}/(x-1)2

Note in the above sum, the k=0 term is 0.

Desired sum=2f'(2)=(n-2)2n+2

(sorry for my mistake!!)
 This is the sum of a Arithmetic-Geometric Series: S_n=n2^n-2^(n+1)+2
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