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Removing summation notation

marobin

New member
Feb 22, 2012
6
I have the following summation and I'm attempting to remove the summation notation. It appears to be the sum of a geometric series but I'm having a great deal of trouble with it. X is an unknown constant.

$$\sum\limits_{i=2}^n (n - (n-i))x^{n-i}$$

Thanks.
 

pickslides

Member
Feb 1, 2012
57
Do you know how to expand the summation out? If so do it for the first 3 or 4 terms, what do you get?
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
I have the following summation and I'm attempting to remove the summation notation. It appears to be the sum of a geometric series but I'm having a great deal of trouble with it. X is an unknown constant.
$$\sum\limits_{i=2}^n (n - (n-i))x^{n-i}$$
Well it reduces to $\sum\limits_{i=2}^n (i)x^{n-i}=2x^{n-2}+3x^{n-3}\cdots+(n-1)x+n$
But that is not very much help.
 

marobin

New member
Feb 22, 2012
6
Thank you for expanding out the first few terms. I am try to find a general formula for it instead of using the summation notation.
 
Jan 31, 2012
54
I have the following summation and I'm attempting to remove the summation notation. It appears to be the sum of a geometric series but I'm having a great deal of trouble with it. X is an unknown constant.

$$\sum\limits_{i=2}^n (n - (n-i))x^{n-i}$$

Thanks.


$$\sum\limits_{i=2}^n (n - (n-i))x^{n-i}=\sum\limits_{i=2}^n ix^{n-i}=x^n\sum\limits_{i=2}^n \frac{i}{x^i}$$


$$\sum\limits_{i=2}^n \frac{i}{x^i}=2(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+...+\frac{1}{x^n})$$

$$+(\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+...+\frac{1}{x^n})+(\frac{1}{x^4}+\frac{1}{x^5}+\frac{1}{x^6}+...+\frac{1}{x^n})+...+\frac{1}{x^n}$$


Could you proceed?
 
Last edited:

marobin

New member
Feb 22, 2012
6
Thank you so much. I understand a bit more now, I guess I'm still a bit confused about

$$\sum\limits_{i=2}^n \frac{i}{x^i}=2(\frac{1}{x^2}+\frac{1}{x^3}+\frac{ 1}{x^4}+\frac{1}{x^5}+...+\frac{1}{x^n})+(\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+...+ \frac{1}{x^n})+(\frac{1}{x^4}+\frac{1}{x^5}+\frac{ 1}{x^6}+...+\frac{1}{x^n})+...+\frac{1}{x^n}$$

in terms of why only the first set of summations is multiplied by 2. Sorry, I'm very rusty on the rules of summation.
 
Jan 31, 2012
54
Thank you so much. I understand a bit more now, I guess I'm still a bit confused about

$$\sum\limits_{i=2}^n \frac{i}{x^i}=2(\frac{1}{x^2}+\frac{1}{x^3}+\frac{ 1}{x^4}+\frac{1}{x^5}+...+\frac{1}{x^n})+(\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+...+ \frac{1}{x^n})+(\frac{1}{x^4}+\frac{1}{x^5}+\frac{ 1}{x^6}+...+\frac{1}{x^n})+...+\frac{1}{x^n}$$

in terms of why only the first set of summations is multiplied by 2. Sorry, I'm very rusty on the rules of summation.

I'm little late to work, so just try to answer my question:

What is $$\sum\limits_{i=2}^n \frac{i}{x^i}$$ ?

$$\sum\limits_{i=2}^n \frac{i}{x^i}=\frac{2}{x^2}+\frac{3}{x^3}+\frac{4}{x^4}+...+\frac{n}{x^n}$$

Yes?

So, why then the above equals to what I wrote in my previous post?


Good-luck! :)
 

marobin

New member
Feb 22, 2012
6
Ah, I see that now. Thank you. So now I understand the common ratio in each distinct set is 1/x so I'm now working on trying to get the proper equation. :)
 

sbhatnagar

Active member
Jan 27, 2012
95
Here is another method (which you may find easy).

Let $\displaystyle S_n=\sum_{i=2}^{n}\frac{i}{x^n}=\frac{2}{x^2}+ \frac{3}{x^3}+\cdots \frac{n}{x^n} \quad (1)$

Multiply both sides by $\frac{1}{x}$:

$\displaystyle \frac{S_n}{x}=\sum_{i=2}^{n}\frac{i}{x^{i+1}}= \frac{2}{x^3}+\frac{3}{x^4}+\cdots \frac{n}{x^{n+1}} \quad (2)$

Subtract (2) from (1):

$$ S_n \left( 1-\frac{1}{x}\right)=\frac{2}{x^2}+ \left( \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+\cdots \frac{1}{x^n}\right)-\frac{n}{x^{n+1}}$$

Can you proceed?
 

marobin

New member
Feb 22, 2012
6
So it seems like this is a Linear-Geometric Series.

By applying the formula and if the lower bound was 1 instead of 2, I believe its closed form would be: $$\frac{-(n+1)(\frac{1}{x})^{n+1}+n(\frac{1}{x})^{n+2}+x}{(x-1)^{2}}$$

I wanted to make sure I'm taking the correct approach before attempting to see how the formula would change when taking into account that the lower bound is 2 and not 1.