# Finding the Closed Form of a Summation for k*z-k: Step-by-Step Solution

• air-man001
In summary: Now, go look up that formula and use it to calculate the sum and take its derivative.In summary, the conversation is about a problem in finding the z-transform in digital signal processing. The person is struggling to find the closed form and LCKurtz gives a hint to start the sum at k = 1 and let z = 1/x. They also mention that the sum is not a geometric series. The conversation ends with LCKurtz and mr Mark44 reminding the person to put in effort and use the formula for the sum of a geometric series to solve the problem.

#### air-man001

hello every one

I have this summation in my book

infinity
$$\sum$$ k*z-k
k=0

my solution:

infinity
$$\sum$$ k*z-k = z-1+2z-2+3z-3+...
k=0

please i want the closed form (step-by-step)

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Here's a hint. The sum might as well start at k = 1. Let z = 1/x

$$\sum_{k=1}^n kx^k = x\sum_{k=1}^n kx^{k-1}$$

What is inside the sum on the right you should recognize as a derivative of a series that you should be able to sum. Can you take it from there?

LCKurtz , i can't understand your post.

I know the closed form of geometric series =$$\frac{FirstTerm} {1 - CommonRatio}$$

CommonRatio = termn / termn-1

but in my problem , the 'CommonRatio' isn't fixed

term4/term3 not equal to term3/term2

air-man001 said:
I know the closed form of geometric series =$$\frac{FirstTerm} {1 - CommonRatio}$$

CommonRatio = termn / termn-1

but in my problem , the 'CommonRatio' isn't fixed

term4/term3 not equal to term3/term2
Which should suggest to you that your series isn't a geometric series and you're heading down the wrong path. LCKurtz has given you an alternative direction, so see what you can do with that.

LCKurtz said:
Here's a hint. The sum might as well start at k = 1. Let z = 1/x

$$\sum_{k=1}^n kx^k = x\sum_{k=1}^n kx^{k-1}$$

What is inside the sum on the right you should recognize as a derivative of a series that you should be able to sum. Can you take it from there?

air-man001 said:
LCKurtz , i can't understand your post.

I know the closed form of geometric series =$$\frac{FirstTerm} {1 - CommonRatio}$$

CommonRatio = termn / termn-1

but in my problem , the 'CommonRatio' isn't fixed

term4/term3 not equal to term3/term2

Do you understand the part of my hint about

$$\sum_{k=1}^n kx^{k-1}$$

being the derivative of something?

LCKurtz

it is derivative of xk
right?

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mr Mark44

I am studying "z-transform" in Digital Signal Processing,

this is not homework, i studying for final Exam tomorrow

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LCKurtz said:
Do you understand the part of my hint about

$$\sum_{k=1}^n kx^{k-1}$$

being the derivative of something?

air-man001 said:
LCKurtz

it is derivative of xk
right?

Yes, so you have a sum of derivatives. That is the same as the derivative of the sum and you should be able to calculate that sum and take its derivative.

thank you mr.LCKurtz

I will try it now

i can't do it!

You have

$$\sum_{k=1}^n kx^k = x\sum_{k=1}^n kx^{k-1}= x\frac{d}{dx} \left (\sum_{k=1}^n x^k\right )$$

You know the formula for the sum of the geometric series. Use it and do the derivative.

I'm sorry i can't do it!   air-man001 said:
I'm sorry i can't do it!   Get out your algebra book and look up the formula for the sum of the first n terms of a geometric series and use it to evaluate the sum.

Then take its derivative. I'm not going to do it for you.

LCKurtz said:
Get out your algebra book and look up the formula for the sum of the first n terms of a geometric series and use it to evaluate the sum.

Then take its derivative. I'm not going to do it for you.

Firstly..Thank you for help me.

secondly.. this is not homework ... the subject in Digital signal processing (finding the z-transform) only.

i will save the issue rather than understanding

air-man001 said:
mr Mark44

I am studying "z-transform" in Digital Signal Processing,

this is not homework, i studying for final Exam tomorrow
Whether it's homework of exam review, we're not going to do your work for you. We'll help you do it and steer you in the right direction if/when you make a mistake, but you have to put in some of the effort.