Finite Sum - Modified Geometric Series

[cepheid]

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?

 

[CRGreathouse]

I think you're right. Check yourself against http://www.research.att.com/~njas/sequences/A036799 if you like (but watch the offset!).

 

[mathman]

Using elemtary calculus I got (n-1)2ⁿ+2

 

[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.

 

[cepheid]

I think you're right. Check yourself against http://www.research.att.com/~njas/sequences/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!

 

[cepheid]

Using elemtary calculus I got (n-1)2ⁿ+2

Exactly what method did you use?

 

[cepheid]

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!

 

[Hurkyl]

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...

 

[cepheid]

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.

 

[mathman]

Exactly what method did you use?

f(x)=sum(0,n-1)x^k=(xⁿ-1)/(x-1)

f'(x)=sum(0,n-1)kx^k-1={(x-1)nx^n-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!)

 

[Eratosthenes1]

This is the sum of a Arithmetic-Geometric Series:

S_n=n2^n-2^(n+1)+2