# Finite Sum - Modified Geometric Series

• cepheid
In summary, the given series S_n can be evaluated using the formula n2^n-2^(n+1)+2 or through the use of the z transform and elementary calculus. Both methods result in the same answer of (n-2)2^n+2. However, it is important to note that the online encyclopedia of integer sequences lists a different formula, which may need to be checked for accuracy.
cepheid
Staff Emeritus
Gold Member
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?

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

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

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.

CRGreathouse said:
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)

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

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

Exactly what method did you use?

John Creighto said:
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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cepheid said:
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...

Hurkyl said:
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.

cepheid said:
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

## What is a finite sum in mathematics?

A finite sum is a mathematical calculation that involves adding a finite number of terms. It is different from an infinite sum, which involves adding an infinite number of terms.

## What is a modified geometric series?

A modified geometric series is a series in which there is a common ratio between each term, but the first term and/or the common ratio may be modified. This is different from a regular geometric series, in which all terms have the same common ratio.

## What is the formula for calculating a finite sum of a modified geometric series?

The formula for calculating the finite sum of a modified geometric series is Sn = a(1-rn)/(1-r), where Sn is the sum of n terms, a is the first term, and r is the common ratio. This formula assumes that the series starts at n = 1.

## What is the difference between a finite sum and an infinite sum?

The main difference between a finite sum and an infinite sum is the number of terms involved. A finite sum has a fixed number of terms, while an infinite sum has an infinite number of terms. This means that a finite sum can be calculated, while an infinite sum may not have a definite value.

## How can modified geometric series be applied in real life?

Modified geometric series can be applied in various real-life situations, such as calculating compound interest on loans or investments, estimating population growth, and determining the total distance traveled by a moving object with a changing velocity. It can also be used in engineering and physics to study exponential decay and growth.

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